# The Unapologetic Mathematician

## Some compact subspaces

Let’s say we have a compact space $X$. A subset $C\subseteq X$ may not be itself compact, but there’s one useful case in which it will be. If $C$ is closed, then $C$ is compact.

Let’s take an open cover $\{F_i\}_{i\in\mathcal{I}}$ of $C$. The sets $F_i$ are open subsets of $C$, but they may not be open as subsets of $X$. But by the definition of the subspace topology, each one must be the intersection of $C$ with an open subset of $X$. Let’s just say that each $F_i$ is an open subset of $X$ to begin with.

Now, we have one more open set floating around. The complement of $C$ is open, since $C$ is closed! So between the collection $\{F_i\}$ and the extra set $X\setminus C$ we’ve got an open cover of $X$. By compactness of $X$, this open cover has a finite subcover. We can throw out $X\setminus C$ from the subcover if it’s in there, and we’re left with a finite open cover of $C$, and so $C$ is compact.

In fact, if we restrict to Hausdorff spaces, $C$ must be closed to be compact. Indeed, we proved that if $C$ is compact and $X$ is Hausdorff then any point $x\in X\setminus C$ can be separated from $C$ by a neighborhood $U\subseteq X\setminus C$. Since there is such an open neighborhood, $x$ must be an interior point of $X\setminus C$. And since $x$ was arbitrary, every point of $X\setminus C$ is an interior point, and so $X\setminus C$ must be open.

Putting these two sides together, we can see that if $X$ is compact Hausdorff, then a subset $C\subseteq X$ is compact exactly when it’s closed.

January 15, 2008

## Compact Spaces

An amazingly useful property for a space $X$ is that it be “compact”. We define this term by saying that if $\{U_i\}_{i\in\mathcal{I}}$ is any collection of open subsets of $X$ indexed by any (possibly infinite) set $\mathcal{I}$ so that their union $\bigcup\limits_{i\in\mathcal{I}}U_i$ is the whole of $X$ — the sexy words are “open cover” — then there is some finite collection of the index set $\mathcal{A}\subseteq\mathcal{I}$ so that the union of this finite number of open sets $\bigcup\limits_{i\in\mathcal{A}}U_i$ still contains all of $X$ — the sexy words are “has a finite subcover”.

So why does this matter? Well, let’s consider a Hausdorff space $X$, a point $x\in X$, and a finite collection of points $A\subseteq X$. Given any point $a\in A$, we can separate $x$ and $a$ by open neighborhoods $x\in U_a$ and $a\in V_a$, precisely because $X$ is Hausdorff. Then we can take the intersection $U=\bigcap\limits_{a\in A}U_a$ and the union $V=\bigcup\limits_{a\in A}V_a$. The set $U$ is a neighborhood of $X$, since it’s a finite intersection of neighborhoods, while the set $V$ is a neighborhood of $A$. These two sets can’t intersect, and so we have separated $x$ and $A$ by neighborhoods.

But what if $A$ is an infinite set? Then the infinite intersection $\bigcap\limits_{a\in A}U_a$ may not be a neighborhood of $x$! Infinite operations sometimes cause problems in topology, but compactness can make them finite. If $A$ is a compact subset of $X$, then we can proceed as before. For each $a\in A$ we have open neighborhoods $x\in U_a$ and $a\in V_a$, and so $A\subseteq\bigcup\limits_{a\in A}V_a$ — the open sets $V_a$ form a cover of $A$. Then compactness tells us that we can pick a finite collection $A'\subseteq A$ so that the union $V=\bigcup\limits_{a\in A'}V_a$ of that finite collection of sets still covers $A$ — we only need a finite number of the $V_a$ to cover $A$. The finite intersection $U=\bigcap\limits_{a\in A'}U_a$ will then be a neighborhood of $x$ which doesn’t touch $V$, and so we can separate any point $x\in X$ and any compact set $A\subseteq X$ by neighborhoods.

As an exercise, do the exact same thing again to show that in a Hausdorff space $X$ we can separate any two compact sets $A\subseteq X$ and $B\subseteq X$ by neighborhoods.

In a sense, this shows that while compact spaces may be infinite, they sometimes behave as nicely as finite sets. This can make a lot of things simpler in the long run. And just like we saw for connectivity, we are often interested in things behaving nicely near a point. We thus define a space to be “locally compact” if every point has a neighborhood which is compact (in the subspace topology).

There’s an equivalent definition in terms of closed sets, which is dual to this one. Let’s say we have a collection $\{F_i\}_{i\in\mathcal{I}}$ of closed subsets of $X$ so that the intersection of any finite collection of the $F_i$ is nonempty. Then I assert that the intersection of all of the $F_i$ will be nonempty as well if $X$ is compact. To see this, assume that the intersection is empty:
$\bigcap\limits_{i\in\mathcal{I}}F_i=\varnothing$
Then the complement of this intersection is all of $X$. We can rewrite this as the union of the complements of the $F_i$:
$X=\bigcup\limits_{i\in\mathcal{I}}F_i^c$
Since we’re assuming $X$ to be compact, we can find some finite subcollection $\mathcal{A}\subseteq\mathcal{I}$ so that
$X=\bigcup\limits_{i\in\mathcal{A}}F_i^c$
which, taking complements again, implies that
$\bigcap\limits_{i\in\mathcal{A}}F_i=\varnothing$
but we assumed that all of the finite intersections were nonempty!

Now turn this around and show that if we assume this “finite intersection property” — that if all finite intersections of a collection of closed sets $F_i$ are nonempty, then the intersection of all the $F_i$ are nonempty — then we can derive the first definition of compactness from it.

January 14, 2008

## Separation Axioms

Now that we have some vocabulary about separation properties down we can talk about properties of spaces as a whole, called the separation axioms.

First off, we say that a space is $T_0$ if every two distinct points can be topologically distinguished. This fails, for example, in the trivial topology on a set $X$ if $X$ has at least two points, because every point has the same collection of neighborhoods — $\mathcal{N}(x)=\{X\}$ for all points $x\in X$. As far as the topology is concerned, all the points are the same. This turns out to be particularly interesting in conjunction with other separation axioms, since we often will have one axiom saying that a property holds for all distinct points, and another saying that the property holds for all topologically distinguishable points. Adding $T_0$ turns the latter version into the former.

Next, we say that a space is $R_0$ if any two topologically distinguishable points are separated. That is, we never have a point $x$ in the closure of the singleton set $\{y\}$ without the point $y$ being in the closure of $\{x\}$. Adding $T_0$ to this condition gives us $T_1$. A $T_1$ space is one in which any two distinct points are not only topologically distinguishable, but separated. In particular, we can see that the singleton set $\{x\}$ is closed, since its closure can’t contain any other points than $x$ itself.

A space is $R_1$ if any two topologically distinguishable points are separated by neighborhoods. If this also holds for any pair of distinct points we say that the space is $T_2$, or “Hausdorff”. This is where most topologists start to feel comfortable, though the topologies that arise in algebraic geometry are usually non-Hausdorff. To a certain extent (well, to me at least) Hausdorff spaces feel a lot more topologically natural and intuitive than non-Hausdorff spaces, and you almost have to try to construct pathological spaces to violate this property. Back in graduate school, some of us adapted the term to apply more generally, as in “That guy Steve is highly non-Hausdorff.”

One interesting and useful property of Hausdorff spaces is that the image of the diagonal map $\Delta:X\rightarrow X\times X$ defined by $\Delta(x)=(x,x)$ is closed. To see this, notice that it means the complement of the image is open. That is, if $(x,y)$ is a pair of points of $X$ with $x\neq y$ then we can find an open neighborhood containing the point $(x,y)$ consisting only of pairs $(z,w)$ with $z\neq w$. In fact, we have a base for the product topology on $X\times X$ consisting of products two open sets in $X$. That is, we can pick our open neighborhood of $(x,y)$ to be the set of all pairs $(z,w)$ with $z\in Z$ and $w\in W$, where $Z$ is an open subset of $X$ containing $x$ and $W$ is an open subset containing $y$. To say that this product doesn’t touch the diagonal means that $Z\cap W=\varnothing$, which is just what it means for $x$ and $y$ to be separated by neighborhoods!

We can strengthen this by asking that any two distinct points are separated by closed neighborhoods. If this holds we say the space is $T_{2\frac{1}{2}}$. There’s no standard name for the weaker version discussing topologically distinguishable points. Stronger still is saying that a space is “completely Hausdorff” or completely $T_2$, which asks that any two distinct points be separated by a function.

A space $X$ is “regular” if given a point $x\in X$ and a closed subset $C\subseteq X$ with $x\notin C$ we can separate $\{x\}$ and $C$ by neighborhoods. This is a bit stronger than being Hausdorff, where we only asked that this hold for two singletons. For regular spaces, we allow one of the two sets we’re separating to be any closed set. If we add on the $T_0$ condition we’re above $T_1$, and so singletons are just special closed sets anyhow, but we’re strictly stronger than regularity now. We call this condition $T_3$.

As for Hausdorff, we say that a space is completely regular if we can actually separate $\{x\}$ and $C$ by a function. If we take a completely regular space and add $T_0$, we say it’s $T_{3\frac{1}{2}}$, or “completely regular Hausdorff”, or “Tychonoff”.

We say a space is “normal” if any two disjoint closed subsets are separated by neighborhoods. In fact, a theorem known as Urysohn’s Lemma tells us that we get for free that they’re separated by a function as well. If we add in $T_1$ (not $T_0$ this time) we say that it is “normal Hausdorff”, or $T_4$.

A space is “completely normal” if any two separated sets are separated by neighborhoods. Adding in $T_1$ we say that the space is “completely normal Hausdorff”, or $T_5$.

Finally, a space is “perfectly normal” if any two disjoint closed sets are precisely separated by a function. Adding $T_1$ makes the space “perfectly normal Hausdorff”, or $T_6$.

The Wikipedia entry here is rather informative, and has a great schematic showing which of the axioms imply which others. Most of these axioms I won’t be using, but it’s good to have them out here in case I need them.

January 11, 2008

## Separation Properties

There’s a whole list of properties of topological spaces that we may want to refer to called the separation axioms. Even when two points are distinct elements of the underlying set of a topological space, we may not be able to tell them apart with topological techniques. Points are separated if we can tell them apart in some way using the topology. Today we’ll discuss various properties of separation, and tomorrow we’ll list some of the more useful separation axioms we can ask that a space satisfy.

First, and weakest, we say that points $x$ and $y$ in a topological space $X$ are “topologically distinguishable” if they don’t have the same collection of neighborhoods — if $\mathcal{N}(x)\neq\mathcal{N}(y)$. Now maybe one of the collections of neighborhoods strictly contains the other: $\mathcal{N}(x)\subseteq\mathcal{N}(y)$. In this case, every neighborhood of $x$ is a neighborhood of $y$. a forteriori it contains a neighborhood of $y$, and thus contains $y$ itself. Thus the point $x$ is in the closure of the set $\{y\}$. This is really close. The points are topologically distinguishable, but still a bit too close for comfort. So we define points to be “separated” if each has a neighborhood the other one doesn’t, or equivalently if neither is in the closure of the other. We can extend this to subsets larger than just points. We say that two subsets $A$ and $B$ are separated if neither one touches the closure of the other. That is, $A\cap\bar{B}=\varnothing$ and $\bar{A}\cap B=\varnothing$.

We can go on and give stronger conditions, saying that two sets are “separated by neighborhoods” if they have disjoint neighborhoods. That is, there are neighborhoods $U$ and $V$ of $A$ and $B$, respectively, and $U\cap V=\varnothing$. Being a neighborhood here means that $U$ contains some open set $S$ which contains $A$ and $V$ contains some open set $T$ which contains $B$, and so the closure of $A$ is contained in the open set, and thus in $U$. Similarly, the closure of $B$ must be contained in $V$.. We see that the closure of $B$ is contained in the complement of $S$, and similarly the closure of $A$ is in the complement of $T$, so neither $A$ nor $B$ can touch the other’s closure. Stronger still is being “separated by closed neighborhoods”, which asks that $U$ and $V$ be disjoint closed neighborhoods. These keep $A$ and $B$ even further apart, since these neighborhoods themselves can’t touch each other’s closures.

The next step up is that sets be “separated by a function” if there is a continuous function $f:X\rightarrow\mathbb{R}$ so that for every point $a\in A$ we have $f(a)=0$, and for every point $b\in B$ we have $f(b)=1$. In this case we can take the closed interval $\left[-1,\frac{1}{3}\right]$ whose preimage must be a closed neighborhood of $A$ by continuity. Similarly we can take the closed interval $\left[\frac{2}{3},2\right]$ whose preimage is a closed neighborhood of $B$. Since these preimages can’t touch each other, we have separated $A$ and $B$ by closed neighborhoods. Stronger still is that $A$ and $B$ are “precisely separated by a function”, which adds the requirement that only points from $A$ go to ${0}$ and only points from $B$ go to $1$.

This list of separation

January 10, 2008

## The Image of a Connected Space

One theorem turns out to be very important when we’re dealing with connected spaces, or even just with a connected component of a space. If $f$ is a continuous map from a connected space $X$ to any topological space $Y$, then the image $f(X)\subseteq Y$ is connected. Similarly, if $X$ is path-connected then its image is path-connected.

The path-connected version is actually more straightforward. Let’s say that we pick points $y_0$ and $y_1$ in $f(X)$. Then there must exist $x_0$ and $x_1$ with $f(x_0)=y_0$ and $f(x_1)=y_1$. By path-connectedness there is a function $g:\left[0,1\right]\rightarrow X$ with $g(0)=x_0$ and $g(1)=x_1$, and so $f(g(0))=y_0$ and $f(g(1))=y_1$. Thus the composite function $g\circ f:\left[0,1\right]\rightarrow Y$ is a path from $y_0$ to $y_1$.

Now for the connected version. Let’s say that $f(X)$ is disconnected. Then we can write it as the disjoint union of two nonempty closed sets $B_1$ and $B_2$ by putting some connected components in the one and some in the other. Taking complements we see that both of these sets are also open. Then we can consider their preimages $f^{-1}(B_1)$ and $f^{-1}(B_2)$, whose union is $X$ since every point in $X$ lands in either $B_1$ or $B_2$.

By the continuity of $f$, each of these preimages is open. Seeing as each is the complement of the other, they must also both be closed. And neither one can be empty because some points in $X$ land in each of $B_1$ and $B_2$. Thus we have a nontrivial clopen set in $X$, contradicting the assumption that it’s connected. Thus the image $f(X)$ must have been connected, as was to be shown.

From this theorem we see that the image of any connected component under a continuous map $f$ must land entirely within a connected component of the range of $f$. For example, any map from a connected space to a totally disconnected space (one where each point is a connected component) must be constant.

When we specialize to real-valued functions, this theorem gets simple. Notice that a connected subset of $\mathbb{R}$ is just an interval. It may contain one or both endpoints, and it may stretch off to infinity in one or both directions, but that’s about all the variation we’ve got. So if $X$ is a connected space then the image $f(X)$ of a continuous function $f:X\rightarrow\mathbb{R}$ is an interval.

An immediate corollary to this fact is the intermediate value theorem. Given a connected space $X$, a continuous real-valued function $f$, and points $x_1,x_2\in X$ with $f(x_1)=a_1$ and $f(x_2)=a_2$ (without loss of generality, $a_1), then for any $b\in\left(a_1,a_2\right)$ there is a $y\in X$ so that $f(y)=b$. That is, a continuous function takes all the values between any two values it takes. In particular, if $X$ is itself an interval in $\mathbb{R}$ we get back the old intermediate value theorem from calculus.

January 3, 2008

## Connectedness

Tied in with the fundamental notion of continuity for studying topology is the notion of connectedness. In fact, once two parts of a space are disconnected, there’s almost no topological influence of one on the other, which should be clear from an intuitive idea of what it might mean for a space to be connected or disconnected. This intuitive notion can be illustrated by considering subspaces of the real plane $\mathbb{R}^2$.

First, just so we’re clear, a subset of the plane is closed if it contains its boundary and open if it contains no boundary points. Here there’s a lot more between open and closed than there is for intervals with just two boundary points. Anyhow, you should be able to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same answer, but it’s interesting to generalize this to find a metric on the product of two arbitrary metric spaces.

Anyhow, back to connectedness. Take a sheet of paper to be your plane, and draw a bunch of blobs on it. Use dotted lines sometimes to say you’re leaving out that section of the blob’s border. Have fun with it.

Now we’ll consider that collection of blobs as a subspace $X\subseteq\mathbb{R}^2$, and thus it inherits the subspace topology. We can take one blob $B$ and throw an open set around it that doesn’t hit any other blobs (draw a dotted curve around the blob that doesn’t touch any other). Thus the blob is an open subset $B\subseteq X$ because it’s the intersection of the open set we drew in the plane and the subspace $X$. But we could also draw a solid curve instead of a dotted one and get a closed set in the plane whose intersection with $X$ is $B$. Thus $B$ is also a closed subset of $X$. Some people like to call such a subset in a topological space “clopen”.

In general, given any topological space we can break it into clopen sets. If the only clopen sets are the whole space and the empty subspace, then we’re done. Otherwise, given a nontrivial clopen subset, its complement must also be clopen (why?), and so we can break it apart into those pieces. We call a space with no nontrivial clopen sets “connected”, and a maximal connected subspace $B$ of a topological space $X$ we call a “connected component”. That is, if we add any other points from $X$ to $B$, it will be disconnected.

An important property of connected spaces is that we cannot divide them into two disjoint nonempty closed subsets. Indeed, if we could then the complement of one closed subset would be the other. It would be open (as the complement of a closed subset) and closed (by assumption) and nontrivial since neither it nor its complement could be empty. Thus we would have a nontrivial clopen subset, contrary to our assumptions.

If we have a bunch of connected spaces, we can take their coproduct — their disjoint union — to get a disconnected space with the original family of spaces as its connected components. Conversely, any space can be broken into its connected components and thus written as a coproduct of connected spaces. In general, morphisms from the coproduct of a collection of objects exactly correspond to collections of morphisms from the objects themselves. Here this tells us that a continuous function from any space is exactly determined by a collection of continuous functions, one for each connected component. So we don’t really lose much at all by just talking about connected spaces and trying to really understand them.

Sometimes we’re just looking near one point or another, like we’ve done for continuity or differentiability of functions on the real line. In this case we don’t really care whether the space is connected in general, but just that it looks like it’s connected near the point we care about. We say that a space is “locally connected” if every point has a connected neighborhood.

Sometimes just being connected isn’t quite strong enough. Take the plane again and mark axes. Then draw the graph of the function defined by $f(x)=\sin\left(\frac{1}{x}\right)$ on the interval $\left(0,1\right]$, and by $f(0)=0$. We call this the “topologist’s sine curve”. It’s connected because any open set we draw containing the wiggly sine bit gets the point $(0,0)$ too. The problem is, we might want to draw paths between points in the space, and we can’t do that here. For two points in the sine part, we just follow the curve, but we can never quite get to or away from the origin. Incidentally, it’s also not locally connected because any small ball around the origin contains a bunch of arcs from the sine part that aren’t connected to each other.

So when we want to draw paths, we ask that a space be “path-connected”. That is, given points $x_0$ and $x_1$ in our space $X$, there is a function $f:\left[0,1\right]\rightarrow X$ with $f(0)=x_0$ and $f(1)=x_1$. Slightly stronger, we might want to require that we can choose this function to be a homeomorphism from the closed interval $\left[0,1\right]$ onto its image in $X$. In this case we say that the space is “arc-connected”.

Arc-connectedness clearly implies path-connectedness, and we’ll see in more detail later that path-connectedness implies connectedness. However, the converses do not hold. The topologist’s sine curve gives a counterexample where connectedness doesn’t imply path-connectedness, and I’ll let you try to find a counterexample for the other converse.

Just like we had local connectedness, we say that a space is locally path- or arc-connected if every point has a neighborhood which is path- or arc-connected. We also have path-components and arc-components defined as for connected components. Unfortunately, we don’t have as nice a characterization as we did before for a space in terms of its path- or arc-components. In the topologist’s sine curve, for example, the wiggly bit and the point at the origin are the two path components, but they aren’t put together with anything so nice as a coproduct.

[UPDATE]: As discussed below in the comments, I made a mistake here, implicitly assuming the same thing edriv said explicitly. As I say below, point-set topology and analysis really live on the edge of validity, and there’s a cottage industry of crafting counterexamples to all sorts of theorems if you weaken the hypotheses just slightly.

January 2, 2008

## The Order on the Real Numbers

We’ve defined the real numbers $\mathbb{R}$ as a topological field by completing the rational numbers $\mathbb{Q}$ as a uniform space, and then extending the field operations to the new points by continuity. Now we extend the order on the rational numbers to make $\mathbb{R}$ into an ordered field.

First off, we can simplify our work greatly by recognizing that we just need to determine the subset $\mathbb{R}^+$ of positive real numbers — those $x\in\mathbb{R}$ with $x\geq0$. Then we can say $x\geq y$ if $x-y\geq0$. Now, each real number is represented by a Cauchy sequence of rational numbers, and so we say $x\geq0$ if $x$ has a representative sequence $x_n$ with each point $x_n\geq 0$.

What we need to check is that the positive numbers are closed under both addition and multiplication. But clearly if we pick $x_n$ and $y_n$ to be nonnegative Cauchy sequences representing $x$ and $y$, respectively, then $x+y$ is represented by $x_n+y_n$ and $xy$ is represented by $x_ny_n$, and these will be nonnegative since $\mathbb{Q}$ is an ordered field.

Now for each $x$, $x-x=0\geq0$, so $x\geq x$. Also, if $x\geq y$ and $y\geq z$, then $x-y\geq0$ and $y-z\geq0$, so $x-z=(x-y)+(y-z)\geq0$, and so $x\geq z$. These show that $\geq$ defines a preorder on $\mathbb{R}$, since it is reflexive and transitive. Further, if $x\geq y$ and $y\geq x$ then $x-y\geq0$ and $y-x\geq0$, so $x-y=0$ and thus $x=y$. This shows that $\geq$ is a partial order. Clearly this order is total because any real number either has a nonnegative representative or it doesn’t.

One thing is a little hazy here. We asserted that if a number and its negative are both greater than or equal to zero, then it must be zero itself. Why is this? Well if $x_n$ is a nonnegative Cauchy sequence representing $x$ then $-x_n$ represents $-x$. Now can we find a nonnegative Cauchy sequence $y_n$ equivalent to $-x_n$? The lowest rational number that $y_n$ can be is, of course, zero, and so $\left|y_n-(-x_n)\right|\geq x_n$. But for $-x_n$ and $y_n$ to be equivalent we must have for each positive rational $r$ an $N$ so that $r\geq\left|y_n-(-x_n)\right|\geq x_n$ for $n\geq N$. But this just says that $x_n$ converges to ${0}$!

So $\mathbb{R}$ is an ordered field, so what does this tell us? First off, we get an absolute value $\left|x\right|$ just like we did for the rationals. Secondly, we’ll get a uniform structure as we do for any ordered group. This uniform topology has a subbase consisting of all the half-infinite intervals $(x,\infty)$ and $(-\infty,x)$ for all real $x$. But this is also a subbase for the metric we got from completing the rationals, and so the two topologies coincide!

One more very important thing holds for all ordered fields. As a field $\mathbb{F}$ is a kind of a ring with unit, and like any ring with unit there is a unique ring homomorphism $\mathbb{Z}\rightarrow\mathbb{F}$. Now since $1gt;0$ in any ordered field, we have $2=1+1>0$, and $3=2+1>0$, and so on, to show that no nonzero integer can become zero under this map. Since we have an injective homomorphism of rings, the universal property of the field of fractions gives us a unique field homomorphism $\mathbb{Q}\rightarrow\mathbb{F}$ extending the ring homomorphism from the integers.

Now if $\mathbb{F}$ is complete in the uniform structure defined by its order, this homomorphism will be uniformly complete. Therefore by the universal property of uniform completions, we will find a unique extension $\mathbb{R}\rightarrow\mathbb{F}$. That is, given any (uniformly) complete ordered field there is a unique uniformly continuous homomorphism of fields from the real numbers to the field in question. Thus $\mathbb{R}$ is the universal such field, which characterizes it uniquely up to isomorphism!

So we can unambiguously speak of “the” real numbers, even if we use a different method of constructing them, or even no method at all. We can work out the rest of the theory of real numbers from these properties (though for the first few we might fall back on our construction) just as we could work out the theory of natural numbers from the Peano axioms.

December 4, 2007

## The Topological Field of Real Numbers

We’ve defined the topological space we call the real number line $\mathbb{R}$ as the completion of the rational numbers $\mathbb{Q}$ as a uniform space. But we want to be able to do things like arithmetic on it. That is, we want to put the structure of a field on this set. And because we’ve also got the structure of a topological space, we want the field operations to be continuous maps. Then we’ll have a topological field, or a “field object” (analogous to a group object) in the category $\mathbf{Top}$ of topological spaces.

Not only do we want the field operations to be continuous, we want them to agree with those on the rational numbers. And since $\mathbb{Q}$ is dense in $\mathbb{R}$ (and similarly $\mathbb{Q}\times\mathbb{Q}$ is dense in $\mathbb{R}\times\mathbb{R}$), we will get unique continuous maps to extend our field operations. In fact the uniqueness is the easy part, due to the following general property of dense subsets.

Consider a topological space $X$ with a dense subset $D\subseteq X$. Then every point $x\in X$ has a sequence $x_n\in D$ with $\lim x_n=x$. Now if $f:X\rightarrow Y$ and $g:X\rightarrow Y$ are two continuous functions which agree for every point in $D$, then they agree for all points in $X$. Indeed, picking a sequence in $D$ converging to $x$ we have
$f(x)=f(\lim x_n)=\lim f(x_n)=\lim g(x_n)=g(\lim x_n)=g(x)$.

So if we can show the existence of a continuous extension of, say, addition of rational numbers to all real numbers, then the extension is unique. In fact, the continuity will be enough to tell us what the extension should look like. Let’s take real numbers $x$ and $y$, and sequences of rational numbers $x_n$ and $y_n$ converging to $x$ and $y$, respectively. We should have
$s(x,y)=s(\lim x_n,\lim y_n)=s(\lim(x_n,y_n))=\lim x_n+y_n$
but how do we know that the limit on the right exists? Well if we can show that the sequence $x_n+y_n$ is a Cauchy sequence of rational numbers, then it must converge because $\mathbb{R}$ is complete.

Given a rational number $r$ we must show that there exists a natural number $N$ so that $\left|(x_m+y_m)-(x_n+y_n)\right| for all $m,n\geq N$. But we know that there’s a number $N_x$ so that $\left|x_m-x_n\right|<\frac{r}{2}$ for $m,n\geq N_x$, and a number $N_y$ so that $\left|y_m-y_n\right|<\frac{r}{2}$ for $m,n\geq N_y$. Then we can choose $N$ to be the larger of $N_x$ and $N_y$ and find
$\left|(x_m-x_n)+(y_m-y_n)\right|\leq\left|x_m-x_n\right|+\left|y_m-y_n\right|<\frac{r}{2}+\frac{r}{2}=r$
So the sequence of sums is Cauchy, and thus converges.

What if we chose different sequences $x'_n$ and $y'_n$ converging to $x$ and $y$? Then we get another Cauchy sequence $x'_n+y'_n$ of rational numbers. To show that addition of real numbers is well-defined, we need to show that it’s equivalent to the sequence $x_n+y_n$. So given a rational number $r$ does there exist an $N$ so that $\left|(x_n+y_n)-(x'_n+y'_n)\right| for all $n\geq N$? This is almost exactly the same as the above argument that each sequence is Cauchy! As such, I’ll leave it to you.

So we’ve got a continuous function taking two real numbers and giving back another one, and which agrees with addition of rational numbers. Does it define an Abelian group? The uniqueness property for functions defined on dense subspaces will come to our rescue! We can write down two functions from $\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$ defined by $s(s(x,y),z)$ and $s(x,s(y,z))$. Since $s$ agrees with addition on rational numbers, and since triples of rational numbers are dense in the set of triples of real numbers, these two functions agree on a dense subset of their domains, and so must be equal. If we take the ${0}$ from $\mathbb{Q}$ as the additive identity we can also verify that it acts as an identity real number addition. We can also find the negative of a real number $x$ by negating each term of a Cauchy sequence converging to $x$, and verify that this behaves as an additive inverse, and we can show this addition to be commutative, all using the same techniques as above. From here we’ll just write $x+y$ for the sum of real numbers $x$ and $y$.

What about the multiplication? Again, we’ll want to choose rational sequences $x_n$ and $y_n$ converging to $x$ and $y$, and define our function by
$m(x,y)=m(\lim x_n,\lim y_n)=m(\lim(x_n,y_n))=\lim x_ny_n$
so it will be continuous and agree with rational number multiplication. Now we must show that for every rational number $r$ there is an $N$ so that $\left|x_my_m-x_ny_n\right| for all $m,n\geq N$. This will be a bit clearer if we start by noting that for each rational $r_x$ there is an $N_x$ so that $\left|x_m-x_n\right| for all $m,n\geq N_x$. In particular, for sufficiently large $n$ we have $\left|x_n\right|<\left|x_N\right|+r_x$, so the sequence $x_n$ is bounded above by some $b_x$. Similarly, given $r_y$ we can pick $N_y$ so that $\left|y_m-y_n\right| for $m,n\geq N_y$ and get an upper bound $b_y\geq y_n$ for all $n$. Then choosing $N$ to be the larger of $N_x$ and $N_y$ we will have
$\left|x_my_m-x_ny_n\right|=\left|(x_m-x_n)y_m+x_n(y_m-y_n)\right|\leq r_xb_y+b_xr_y$
for $m,n\geq N$. Now given a rational $r$ we can (with a little work) find $r_x$ and $r_y$ so that the expression on the right will be less than $r$, and so the sequence is Cauchy, as desired.

Then, as for addition, it turns out that a similar proof will show that this definition doesn’t depend on the choice of sequences converging to $x$ and $y$, so we get a multiplication. Again, we can use the density of the rational numbers to show that it’s associative and commutative, that $1\in\mathbb{Q}$ serves as its unit, and that multiplication distributes over addition. We’ll just write $xy$ for the product of real numbers $x$ and $y$ from here on.

To show that $\mathbb{R}$ is a field we need a multiplicative inverse for each nonzero real number. That is, for each Cauchy sequence of rational numbers $x_n$ that doesn’t converge to ${0}$, we would like to consider the sequence $\frac{1}{x_n}$, but some of the $x_n$ might equal zero and thus throw us off. However, there can only be a finite number of zeroes in the sequence or else ${0}$ would be an accumulation point of the sequence and it would either converge to ${0}$ or fail to be Cauchy. So we can just change each of those to some nonzero rational number without breaking the Cauchy property or changing the real number it converges to. Then another argument similar to that for multiplication shows that this defines a function from the nonzero reals to themselves which acts as a multiplicative inverse.

December 3, 2007

## Complete Uniform Spaces

Okay, in a uniform space we have these things called “Cauchy nets”, which are ones where the points of the net are getting closer and closer to each other. If our space is sequential — usually a result of assuming it to be first- or second-countable — then we can forget the more complicated nets and just consider Cauchy sequences. In fact, let’s talk as if we’re looking at a sequence to build up an intuition here.

Okay, so a sequence is Cauchy if no matter what entourage we pick to give a scale of closeness, there’s some point along our sequence where all of the remaining points are at least that close to each other. If we pick a smaller entourage we might have to walk further out the sequence, but eventually every point will be at least that close to all the points beyond it. So clearly they’re all getting pressed together towards a limit, right?

Unfortunately, no. And we have an example at hand of where it can go horribly, horribly wrong. The rational numbers $\mathbb{Q}$ are an ordered topological group, and so they have a uniform structure. We can give a base for this topology consisting of all the rays $(a,\infty)=\{x\in\mathbb{Q}|a, the rays $(-\infty,a)=\{x\in\mathbb{Q}|x, and the intervals $(a,b)=\{x\in\mathbb{Q}|a, which is clearly countable and thus makes $\mathbb{Q}$ second-countable, and thus sequential.

Okay, I’ll take part of that back. This is only “clear” if you know a few things about cardinalities which I’d thought I’d mentioned but it turns out I haven’t. It was also pointed out that I never said how to generate an equivalence relation from a simpler relation in a comment earlier. I’ll wrap up those loose ends shortly, probably tomorrow.

Back to the business at hand: we can now just consider Cauchy sequences, instead of more general Cauchy nets. Also we can explicitly give entourages that comprise a base for the uniform structure, which is all we really need to check the Cauchy condition: $E_a=\{(x,y)\in\mathbb{Q}\times\mathbb{Q}|\left|x-y\right|. I did do absolute values, didn’t I? So a sequence $x_i$ is Cauchy if for every rational number $a$ there is an index $N$ so that for all $i\geq N$ and $j\geq N$ we have $\left|x_i-x_j\right|.

We also have a neighborhood base $\mathcal{B}(q)$ for each rational number $q$ given by the basic entourages. For each rational number $r$ we have the neighborhood $\{x\in\mathbb{Q}|\left|x-q\right|. These are all we need to check convergence. That is, a sequence $x_i$ of rational numbers converges to $q$ if for all rational $r$ there is an index $N$ so that for all $i\geq N$ we have $\left|x_i-q\right|.

And finally: for each natural number $n\in\mathbb{N}$ there are only finitely many square numbers less than $2n^2$. We’ll let $a_n^2$ be the largest such number, and consider the rational number $x_n=\frac{a_n}{n}$. We can show that this sequence is Cauchy, but it cannot converge to any rational number. In fact, if we had such a thing this sequence would be trying to converge to the square root of two.

The uniform space $\mathbb{Q}$ is shot through with holes like this, making tons of examples of Cauchy sequences which “should” converge, but don’t. And this is all just in one little uniform space! Clearly Cauchy nets don’t converge in general. But we dearly want them to. If we have a uniform space in which every Cauchy sequence does converge, we call it “complete”.

Categorically, a complete uniform space is sort of alike an abelian group. The additional assumption is an extra property which we may forget when convenient. That is, we have a category $\mathbf{Unif}$ of uniform spaces and a full subcategory $\mathbf{CUnif}$ of complete uniform spaces. The inclusion functor of the subcategory is our forgetful functor, and we’d like an adjoint to this functor which assigns to each uniform space $X$ its “completion” $\overline{X}$. This will contain $X$ as a dense subspace — the closure $\mathrm{Cl}(X)$ in $\overline{X}$ is the whole of $\overline{X}$ — and will satisfy the universal property that if $Y$ is any other complete uniform space and $f:X\rightarrow Y$ is a uniformly continuous map, then there is a unique uniformly continuous $\bar{f}:\overline{X}\rightarrow Y$ extending $f$.

To construct such a completion, we’ll throw in the additional assumption that $X$ is second-countable so that we only have to consider Cauchy sequences. This isn’t strictly necessary, but it’s convenient and gets the major ideas across. I’ll leave you to extend the construction to more general uniform spaces if you’re interested.

What we want to do is identify Cauchy sequences in $X$ — those which should converge to something in the completion — with their limit points in the completion. But more than one sequence might be trying to converge to the same point, so we can’t just take all Cauchy sequences as points. So how do we pick out which Cauchy sequences should correspond to the same point? We’ll get at this by defining what the uniform structure (and thus the topology) should be, and then see which points have the same neighborhoods.

Given an entourage $E$ of $X$ we can define an entourage $\overline{E}$ as the set of those pairs of sequences $(x_i,y_j)$ where there exists some $N$ so that for all $i\geq N$ and $j\geq N$ we have $(x_i,y_j)\in E$. That is, the sequences which get eventually $E$-close to each other are considered $\overline{E}$-close.

Now two sequences will be equivalent if they are $\overline{E}$-close for all entourages $E$ of $X$. We can identify these sequences and define the points of $\overline{X}$ to be these equivalence classes of Cauchy sequences. The entourages $\overline{E}$ descend to define entourages on $\overline{X}$, thus defining it as a uniform space. It contains $X$ as a uniform subspace if we identify $x\in X$ with (the equivalence class of) the constant sequence $x, x, x, ...$. It’s straightforward to show that this inclusion map is uniformly continuous. We can also verify that the second-countability of $X$ lifts up to $\overline{X}$.

Now it also turns out that $\overline{X}$ is complete. Let’s consider a sequence of Cauchy sequences $(x_k)_i$. This will be Cauchy if for all entourages $\overline{E}$ there is an $\bar{N}$ so that if $i\geq\bar{N}$ and $j\geq\bar{N}$ the pair $((x_k)_i,(x_k)_j)$ is in $\overline{E}$. That is, there is an $N_{i,j}$ so that for $k\geq N_{i,j}$ and $l\geq N_{i,j}$ we have $((x_k)_i,(x_l)_j)\in E$. We can’t take the limits in $X$ of the individual Cauchy sequences $(x_k)_i$ — the limits along $k$ — but we can take the limits along $i$! This will give us another Cauchy sequence, which will then give a limit point in $\overline{X}$.

As for the universal property, consider a uniformly continuous map $f:X\rightarrow Y$ to a complete uniform space $Y$. Then every point $\bar{x}$ in $\overline{X}$ comes from a Cauchy sequence $x_i$ in $X$. Being uniformly continuous, $f$ will send this to a Cauchy sequence $f(x_i)$ in $Y$, which must then converge to some limit $\bar{f}(\bar{x})\in Y$ since $Y$ is complete. On the other hand, if $x_i'$ is another representative of $\bar{x}$ then the uniform continuity of $f$ will force $\lim f(x_i)=\lim f(x_i')$, so $\bar{f}$ is well-defined. It is unique because there can be only one continuous function on $\overline{X}$ which agrees with $f$ on the dense subspace $X$.

So what happens when we apply this construction to the rational numbers $\mathbb{Q}$ in an attempt to patch up all those holes and make all the Cauchy sequences converge? At long last we have the real numbers $\mathbb{R}$! Or, at least, we have the underlying complete uniform space. What we don’t have is any of the field properties we’ll want for the real numbers, but we’re getting close to what every freshman in calculus thinks they understand.

November 29, 2007

## Countability Axioms

Now I want to toss out a few assumptions that, if they happen to hold for a topological space, will often simplify our work. There are a lot of these, and the ones that I’ll mention I’ll dole out in small, related collections. Often we will impose one of these assumptions and then just work in the subcategory of $\mathbf{Top}$ of spaces satisfying them, so I’ll also say a few things about how these subcategories behave. Often this restriction to “nice” spaces will end up breaking some “nice” properties about $\mathbf{Top}$, and Grothendieck tells us that it’s often better to have a nice category with some bad objects than to have a bad category with only nice objects. Still, the restrictions can come in handy.

First I have to toss out the concept of a neighborhood base, which is for a neighborhood filter like a base for a topology. That is, a collection $\mathcal{B}(x)\subseteq\mathcal{N}(x)$ of neighborhoods of a point $x$ is a base for the neighborhood filter $\mathcal{N}(x)$ if for every neighborhood $N\in\mathcal{N}(x)$ there is some neighborhood $B\in\mathcal{B}(x)$ with $B\subseteq N$. Just like we saw for a base of a topology, we only need to check the definition of continuity at a point $x$ on a neighborhood base at $x$.

Now we’ll say that a topological space is “first-countable” if each neighborhood filter has a countable base. That is, the sets in $\mathcal{B}(x)$ can be put into one-to-one correspondence with some subset of the natural numbers $\mathbb{N}$. We can take this collection of sets in the order given by the natural numbers: $B_i$. Then we can define $U_0=B_0$, $U_1=U_0\cap B_1$, and in general $U_n=U_{n-1}\cap B_n$. This collection $U_i$ will also be a countable base for the neighborhood filter, and it satisfies the extra property that $m\geq n$ implies that $U_m\subseteq U_n$. From this point we will assume that our countable base is ordered like this.

Why does it simplify our lives to only have a countable neighborhood base at each point? One great fact is that a function $f:X\rightarrow Y$ from a first-countable space $X$ will be continuous at $x$ if each neighborhood $V\in\mathcal{N}_Y(f(x))$ contains the image of some neighborhood $U\in\mathcal{N}_X(x)$. But $U$ must contain a set from our countable base, so we can just ask if there is an $i\in\mathbb{N}$ with $f(B_i)\in V$.

We also picked the $B_i$ to nest inside of each other. Why? Well we know that if $f$ isn’t continuous at $x$ then we can construct a net $x_\alpha\in X$ that converges to $x$ but whose image doesn’t converge to $f(x)$. But if we examine our proof of this fact, we can look only at the base $B_i$ and construct a sequence that converges to $x$ and whose image fails to converge to $f(x)$. That is, a function from a first-countable space is continuous if and only if $\lim f(x_i)=f(\lim x_i)$ for all sequences $x_i\in X$, and sequences are a lot more intuitive than general nets. When this happens we say that a space is “sequential”, and so we have shown that every first-countable space is sequential.

Every subspace of a first-countable space is first-countable, as is every countable product. Thus the subcategory of $\mathbf{Top}$ consisting of first-countable spaces has all countable limits, or is “countably complete”. Disjoint unions of first-countable spaces are also first-countable, so we still have coproducts, but quotients of first-countable spaces may only be sequential. On the other hand, there are sequential spaces which are not first-countable whose subspaces are not even sequential, so we can’t just pass to the subcategory of sequential spaces to recover colimits.

A stronger condition than first-countability is second-countability. This says that not only does every neighborhood filter have a countable base, but that there is a countable base for the topology as a whole. Clearly given any point $x$ we can take the sets in our base which contain $x$ and thus get a countable neighborhood base at that point, so any second-countable space is also first-countable, and thus sequential.

Another nice thing about second-countable spaces is that they are “separable”. That is, in a second-countable space $X$ there will be a countable subset $S\subseteq X$ whose closure $\mathrm{Cl}(S)$ is all of $X$. That is, given any point $x\in X$ there is a sequence $x_i\in S$ — we don’t need nets because $X$ is sequential — so that $x_i$ converges to $x$. That is, in some sense we can “approximate” points of $X$ by sequences of points in $S$, and $S$ itself has only countably many points.

The subcategory of all second-countable spaces is again countably complete, since subspaces and countable products of second-countable spaces are again second-countable. Again, we have coproducts, but not coequalizers since a quotient of a second-countable space may not be second-countable. However, if the map $X\rightarrow X/\sim$ sends open sets in $X$ to open sets in the quotient, then the quotient space is second-countable, so that’s not quite as bad as first-countability.

Second-countability (and sometimes first-countability) is a property that makes a number of constructions work out a lot more easily, and which doesn’t really break too much. It’s a very common assumption since pretty much every space an algebraic topologist or a differential geometer will think of is second-countable. However, as is usually the case with such things, “most” spaces are not second-countable. Still, it’s a common enough assumption that we will usually take it as read, making explicit those times when we don’t assume that a space is second-countable.

November 28, 2007