# The Unapologetic Mathematician

## The Heine-Borel Theorem

We’ve talked about compact subspaces, particularly of compact spaces and Hausdorff spaces (and, of course, compact Hausdorff spaces). So how can we use this to understand the space $\mathbb{R}$ of real numbers, or higher-dimensional versions like $\mathbb{R}^n$?

First off, $\mathbb{R}$ is Hausdorff, which should be straightforward to prove. Unfortunately, it’s not compact. To see this, consider the open sets of the form $(-x,x)$ for all positive real numbers $x$. Given any real number $y$ we can find an $x$ with $|y|, so $y\in(-x,x)$. Therefore the collection of these open intervals covers $\mathbb{R}$. But if we take any finite number of them, one will be the biggest, and so we must miss some real numbers. This open cover does not have a finite subcover, and $\mathbb{R}$ is not compact. We can similarly show that $\mathbb{R}^n$ is Hausdorff, but not compact.

So, since $\mathbb{R}^n$ is Hausdorff, any compact subset of $\mathbb{R}^n$ must be closed. But not every closed subset is compact. What else does compactness imply? Well, we can take the proof that $\mathbb{R}^n$ isn’t compact and adapt it to any subset $A\subseteq\mathbb{R}^n$. We take the collection of all open “cubes” $(-x,x)^n$ consisting of $n$-tuples of real numbers, each of which is between $-x$ and $x$, and we form open subsets of $A$ by the intersections $U_x=(-x,x)^n\cap A$. Now the only way for there to be a finite subcover of this open cover of $A$ is for there to be some $x$ so that $U_x=A$. That is, every component of every point of $A$ has absolute value less than $x$, and so we say that $A$ is “bounded”.

We see now that every compact subset of $\mathbb{R}^n$ is closed and bounded. It turns out that being closed and bounded is not only necessary for compactness, but they’re also sufficient! To see this, we’ll show that the closed cube $\left[-x,x\right]^n$ is compact. Then a bounded set $A$ is contained in some such cube, and a closed subset of a compact space is compact. This is the Heine-Borel theorem.

In the $n=1$ case, we just need to see that the interval $\left[-x,x\right]$ is compact. Take an open cover $\{U_i\}$ of this interval, and define the set $S$ to consist of all $y\in\left[-x,x\right]$ so that a finite collection of the $U_i$ cover $\left[-x,y\right]$. Then define $t$ to be the least upper bound of $S$. Basically, $t$ is as far along the interval as we can get with a finite number of sets, and we’re hoping to show that $t=x$. Clearly it can’t go past $x$, since $S\subseteq\left[-x,x\right]$. But can it be less than $x$?

In fact it can’t, because if it were, then we can find some open set $U$ from the cover that contains $t$. As an open neighborhood of $t$, the set $U$ contains some interval $(t-\epsilon,t+\epsilon)$. Then $t-\epsilon$ must be in $S$, and so there is some finite collection of the $U_i$ which covers $\left[-x,t-\epsilon\right]$. But then we can just add in $U$ to get a finite collection of the $U_i$ which covers $\left[-x,t+\frac{\epsilon}{2}\right]$, and this contradicts the fact that $t$ is the supremum of $S$. Thus $t=x$ and there is a finite subcover of $\left[-x,x\right]$, making this closed interval compact!

Now Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube $\left[-x,x\right]^n\subseteq\mathbb{R}^n$ is compact. And since any closed and bounded set is contained in some such cube, it will be compact as a closed subspace of a compact space. Incidentally, since $n$ is finite, we don’t need to wave the Zorn talisman to get this invocation of the Tychonoff magic to work.

As a special case, we can look back at the one-dimensional case to see that a compact, connected space must be a closed interval $\left[a,b\right]$. Then we know that the image of a connected space is connected, and that the image of a compact space is compact, so the image of a closed interval under a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is another closed interval.

The fact that this image is an interval gave us the intermediate value theorem. The fact that it’s closed now gives us the extreme value theorem: a continuous, real-valued function $f$ on a closed interval $\left[a,b\right]$ attains a maximum and a minimum. That is, there is some $c\in\left[a,b\right]$ so that $f(c)\geq f(x)$ for all $x\in\left[a,b\right]$, and similarly there is some $d\in\left[a,b\right]$ so that $f(d)\leq f(x)$ for all $x\in\left[a,b\right]$.

January 18, 2008

## Tychonoff’s Theorem

One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family $\{X_i\}_{i\in\mathcal{I}}$ of compact spaces is again compact. Unsurprisingly, the really tough bit comes in when we look at an infinite product. Our approach will use the dual definition of compactness.

Let’s say that a collection $\mathcal{F}$ of closed sets has the finite intersection hypothesis if all finite intersections of members of the collection are nonempty, so compactness says that any collection satisfying the finite intersection hypothesis has nonempty intersection. We can then form the collection $\Omega=\{\mathcal{F}\}$ of all collections of sets satisfying the finite intersection hypothesis. This can be partially ordered by containment — $\mathcal{F}'\leq\mathcal{F}$ if every set in $\mathcal{F}'$ is also in $\mathcal{F}$.

Given any particular collection $\mathcal{F}$ we can find a maximal collection containing it by finding the longest increasing chain in $\Omega$ starting at $\mathcal{F}$. Then we simply take the union of all these collections to find the collection at its top. This is almost exactly the same thing as we did back when we showed that every vector space is a free module! And just like then, we need Zorn’s lemma to tell us that we can manage the trick in general, but if we look closely at how we’re going to use it we’ll see that we can get away without Zorn’s lemma for finite products.

Anyhow, this maximal collection $\mathcal{F}$ has two nice properties: it contains all of its own finite intersections, and it contains any set which intersects each set in $\mathcal{F}$. These are both true because if $\mathcal{F}$ didn’t contain one of these sets we could throw it in, make $\mathcal{F}$ strictly larger, and still satisfy the finite intersection hypothesis.

Now let’s assume that $\mathcal{F}$ is a collection of closed subsets of $\prod\limits_{i\in\mathcal{I}}X_i$ satisfying the finite intersection hypothesis. We can then get a maximal collection $\mathcal{G}$ containing $\mathcal{F}$. Then given an index $i\in\mathcal{I}$ we can consider the collection $\{\overline{\pi_i(G)}\}_{G\in\mathcal{G}}$ of closed subsets of $X_i$ and see that it, too, satisfies the finite intersection hypothesis. Thus by compactness of $X_i$ the intersection of this collection is nonempty. Letting $U_i$ be a closed set containing one of these intersection points $x_i$, we see that the preimage $\pi_i^{-1}(U_i)$ meets every $G\in\mathcal{G}$, and so must itself be in $\mathcal{G}$.

Okay, so let’s take the point $x_i$ for each index and consider the point $p$ in $\prod\limits_{i\in\mathcal{I}}X_i$ with $i$-th coordinate $x_i$. Then pick some set $V=\prod\limits_{i\in\mathcal{I}}V_i$ containing $p$ from the base for the product topology. For all but a finite number of the $i$, $V_i=X_i$. For those finite number where it’s smaller, the closure of $V_i$ contains the point $x_i\in X_i$, and so $\pi_i^{-1}(V_i)$ is in $\mathcal{G}$. So their finite intersection must be nonempty, and so is $V$ itself!

Now, since $V$ is in $\mathcal{G}$, it must intersect each of the closed sets in the original collection $\mathcal{F}$. Since the only constraint on $V$ is that it contain $p$, this point must be a limit point of each of the sets in $\mathcal{F}$. And because they’re closed, they must contain all of their limit points. Thus the intersection of all the sets in $\mathcal{F}$ is nonempty, and the product space is compact!

January 17, 2008

## The Image of a Compact Space

One of the nice things about connectedness is that it’s preserved under continuous maps. It turns out that compactness is the same way — the image of a compact space $X$ under a continuous map $f:X\rightarrow Y$ is compact.

Let’s take an open cover $\{U_i\}$ of the image $f(X)$. Since $f$ is continuous, we can take the preimage of each of these open sets $\{f^{-1}(U_i)\}$ to get a bunch of open sets in $X$. Clearly every point of $X$ is the preimage of some point of $f(X)$, so the $f^{-1}(U_i)$ form an open cover of $X$. Then we can take a finite subcover by compactness of $X$, picking out some finite collection of indices. Then looking back at the $U_i$ corresponding to these indices (instead of their preimages) we get a finite subcover of $f(X)$. Thus any open cover of the image has a finite subcover, and the image is compact.

January 16, 2008

## 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

## Algebraic Geometry Blogging

I’m still bleary from a 17 hour drive, so I’m punting.

Over at Rigorous Trivialities, Charles has been working for a few weeks on a series of posts about algebraic geometry “from the beginning”. The expository style seems based on a “just-in-time” delivery model, introducing concepts right before they’re needed. Personally, I always worry that this will lead to way too much backtracking when I realize I forgot to mention a whole line of reasoning that I’ve internalized and don’t consciously think about anymore, but which isn’t trivial. But he seems to be pulling this off well.

I know a bit about algebraic geometry myself — my advisor was fond of saying he remembers it having something to do with solving polynomials — and I’ll get to what I know eventually. I’m sure I’ll be referring to Charles’ notes when I do.

January 9, 2008 Posted by | Algebraic Geometry | 2 Comments

## The Ham Sandwich Theorem

I’ve been up all night so I can sleep early tonight, wake up really early tomorrow, and hit the road back to New Orleans. As a result, I’m really not up for thinking too hard right now, but I probably won’t even get a chance to post tomorrow, so I’ll take a cute from a commenter on my Intermediate Value Theorem post and mention the “ham sandwich” theorem.

The MathWorld entry is pretty good for references here. Basically, if we have three solids (whatever those are) floating out anywhere in three-dimensional space, like two slices of bread and a slice of ham, then you can cut them all in half with one plane. Or if you have four four-dimensional solids you can cut them all in half by a three-dimensional hyperplane. And so on.

The sketch of the proof on that page is pretty clear, I think. You pick a direction and consider planes perpendicular to it. The IVT gives you one plane for each of the three solids. Then you use these to construct a map from the sphere to the plane whose image you can show must contain the point $(0,0)$, which corresponds to all three planes lining up in that direction.

Interestingly, the last step also boils down to something like saying that the image of a connected space under a continuous map is connected. But it’s not quite that. We’ll deal with it later, rest assured.