# The Unapologetic Mathematician

## Neighborhoods

As we go on, we’re going to want to focus right in on a point $x$ in a topological space $(X,\tau)$. We’re interested in the subsets of $X$ in which we could “wiggle $x$ around a bit” without leaving the subset. These subsets we’ll call “neighborhoods” of $x$.

More formally, a neighborhood $N$ of $x$ is a subset of $X$ which contains some open subset $U\subseteq X$, which contains the point $x$. Then we can wiggle $x$ within the “nearby” set $U$ and not leave $N$. Notice here that I’m not requiring $N$ itself to be open, though if it is we call it an “open neighborhood” of $x$.

In fact, any open set $U$ is an open neighborhood of any of its points $x$, since clearly it contains an open set containing $x$ — itself! Similarly, a subset is a neighborhood of any point in its interior. But what about a point not in its interior? If we take a point $x\in S$, but $x\notin S^\circ$, then $S$ is a neighborhood of $X$ if and only if there is an open set $U$ with $x\in U\subseteq S$. But then since $U$ is an open set contained in $S$, we must have $U\subseteq S^\circ$, which would put $x$ into the interior as well. That is, a set is a neighborhood of exactly those points in its interior. In fact, some authors use this condition to define “interior” rather than the one more connected to orders.

So the only way for a set to be a neighborhood of all its points is for all of its points to be in its interior. That is, $S^\circ=S$. But, dually to the situation for the closure operator, the fixed points of the interior operator are exactly the open sets. And so we conclude that a set $S$ is open if and only if it is a neighborhood of all its points — another route to topologies! We say what the neighborhoods of each point are, and then we define an open set as one which is a neighborhood of each of its points.

But now we have to step back a moment. I can’t just toss out any collection of sets and declare them to be the neighborhoods of $x$. There are certain properties that the collection of neighborhoods of a given point must satisfy, and only when we satisfy them will we be able to define a topology in this way. Let’s call something which satisfies these conditions (which we’ll work out) a “neighborhood system” for $x$ and write it $\mathcal{N}(x)$.

First of all (and almost trivially), each set in $\mathcal{N}(x)$ must contain $x$. We’re not going to get much of anywhere if we don’t at least require that.

If $S\in\mathcal{N}(x)$ is a neighborhood of $x$, and $S\subseteq T$, then $T$ must be a neighborhood as well since it contains whatever open set $U$ satisfies the neighborhood condition $x\in U\subseteq S$. Also, if $S$ and $T$ are two neighborhoods of $x$ then $x\in S^\circ\cap T^\circ=(S\cap T)^\circ\subseteq S\cap T$. That is, there must be a neighborhood contained in both $S$ and $T$. We can sum up these two conditions by saying that $\mathcal{N}(x)$ is a “filter” in the partially-ordered set $P(X)$.

So, given a topology $\tau$ on $X$ we get a filter $\mathcal{N}(x)$ for each point. Conversely, if we have such a choice of a filter at each point, we can declare the open sets to be those $U$ so that $x\in U$ implies that $U\in\mathcal{N}(x)$.

Trivially, $\varnothing$ satisfies this condition, as it doesn’t have any points to be a neighborhood of. The total space $X$ satisfies this condition because it’s above everything, so it’s in every filter, and thus is a neighborhood of every point.

Now let’s take two sets $U$ and $V$, which are neighborhoods of each of their points, and let’s consider their intersection and a point $x\in U\cap V$. Since $x$ is in both $U$ and $V$, each of them is a neighborhood of $x$, and so since $\mathcal{N}(x)$ is a filter we see that $U\cap V$ must be a neighborhood of $x$ as well. We can extend this to cover all finite intersections.

On the other hand, let’s consider an arbitrary family $U_\alpha$ of sets, each of which is a neighborhood of each of its points. Now, given any point $x$ in the union $\bigcup\limits_\alpha U_\alpha$ it must be in at least one of the sets, say $U_A$. Now $x\in U_A\subseteq\bigcup\limits_\alpha U_\alpha$ tells us first that $U_A\in\mathcal{N}(x)$ by assumption, and then that $\bigcup\limits_\alpha U_\alpha\in\mathcal{N}(x)$ by the filter property of $\mathcal{N}(x)$. Thus we can take arbitrary intersections, and so we have a topology.

One caveat here: I might be missing something. Other definitions of a neighborhood system tend to include something along the lines of saying that every neighborhood of a point $x$ contains another neighborhood in its interior. I seem to have come up with a topology without using that assumption, but I’m willing to believe that there’s something I’ve missed here. If you see it, go ahead and let me know.

November 15, 2007

## Faulhaber’s Fabulous Formula

Tonight I’ll be telling the undergrad math club here about a nifty little thing I picked up from Scott Carter last week, who in turn picked it up from John Conway shortly before that: Faulhaber’s Fabulous Formula. The formula itself expresses the sum of the first $n$ $p$-th powers as (sort of) an integral:

$\displaystyle\sum\limits_{k=1}^nk^p=\int\limits_b^{b+n}x^p\,dx$

The question here is what this mysterious $b$ is, and the answer to that takes us into the shadowy netherworld of 19th-century analysis: the umbral calculus.

Our mysterious character $b$ is defined by a simple equation: $b^n=(b-1)^n$. For $n=0$ this is trivially true, as both sides are $1$. For $n=1$ this is clearly nonsense, as it says $b^1=b^1-1$, or $0=-1$, so let’s just not apply the relation for this value of $n$. For higher $n$, it’s still nonsense, but of a subtler sort. So let’s dive in.

When we consider $n=2$ we get $b^2=b^2-2b^1+1$. The $b^2$ terms cancel, and what’s left tells us that $b^1=\frac{1}{2}$. So $b=\frac{1}{2}$? Not quite, because that would make sense!

Let’s look at $n=3$. Now the relation reads $b^3=b^3-3b^2+3b^1-1$. Again the $b^3$ terms cancel, and now we know that $b^1=\frac{1}{2}$. Thus we can solve to find $b^2=1/6$. Huh?

Moving on to $n=4$, we find $b^4=b^4-4b^3+6b^2-4b^1+1$. The $b^4$ terms cancel and we substitute in the values we know for $b^2$ and $b^1$ to find $b^3=0$. As I told Scott, “you do realize that this is f___ing nuts, right?”

We can keep going on like this, spinning out values of $b^n$ for all $n\geq1$$b^4=-\frac{1}{30}$, $b^5=0$, $b^6=\frac{1}{42}$, $b^7=0$, $b^8=-\frac{1}{30}$, $b^9=0$ — which have absolutely nothing to do with powers of any base $b$. But we use them as powers in the relation. Remember: this is prima facie nonsense.

But onwards, mathematical soldiers, integrating as to war: we take the integral and break it up into $n$ pieces:

$\displaystyle\int\limits_b^{b+n}x^p\,dx=\sum\limits_{k=1}^n\int\limits_{b+k-1}^{b+k}x^p\,dx$

Even (most of) my calculus students can take the antiderivatives on the right to get

$\displaystyle\frac{x^{p+1}}{p+1}\Biggl\vert_{b-1+k}^{b+k}=\frac{(b+k)^{p+1}-((b-1)+k)^{p+1}}{p+1}$

Now we use the binomial formula to expand each of the terms in the numerator.

$\displaystyle\frac{1}{p+1}\left(\sum\limits_{l=0}^{p+1}\binom{p+1}{l}b^lk^{p+1-l}-\sum\limits_{l=0}^{p+1}\binom{p+1}{l}(b-1)^lk^{p+1-l}\right)$

and combine to find

$\displaystyle\frac{1}{p+1}\sum\limits_{l=0}^{p+1}\binom{p+1}{l}k^{p+1-l}\left(b^l-(b-1)^l\right)$

But we said that $b^n=(b-1)^n$ unless $n=1$! That is, all these terms cancel except for one, which is just $k^p$. And so the integral we started with is the sum of the first $n$ $p$-th powers, as we wanted to show.

So how do we apply this? Well, we can evaluate the integral more directly rather than breaking it into a big sum.

$\displaystyle\int_b^{b+n}x^p\,dx=\frac{(b+n)^{p+1}-b^{p+1}}{p+1}=\sum\limits_{l=0}^{p}\binom{p+1}{l}\frac{n^{p+1-l}}{p+1}b^l$

And now we use the values we cranked out before. Let’s write it out for $p=1$, which many of you already know the answer for.

$\displaystyle\binom{2}{0}\frac{n^2}{2}b^0+\binom{2}{1}\frac{n}{2}b^1=\frac{n^2}{2}+\frac{n}{2}=\frac{n(n+1)}{2}$

What about $p=2$? It’s not as well-known, but some of you might recognize it when we write it out.

$\displaystyle\binom{3}{0}\frac{n^3}{3}b^0+\binom{3}{1}\frac{n^2}{3}b^1+\binom{3}{2}\frac{n}{3}b^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\frac{n(n+1)(2n+1)}{6}$

In fact, given the numbers we cranked out back at the start of the post, we can use this formula to easily write out the sum of the first $n$ 9th powers:

$\displaystyle\frac{n^2(n+1)^2(n^2+n-1)(2n^4+4n^3-n^2-3n+3)}{20}$

all thanks to Faulhaber’s Fabulous Formula.

November 14, 2007 Posted by | Uncategorized | 10 Comments

## Interiors and Closures

When we pick a topology $\tau$ on a set $X$, not every subset is open, and not every subset is closed. However, we can still come up with some open and closed sets from any subset $U$.

For the open set, notice that we always have at least one open set inside $U$: the empty set. So we can gather up all the open sets contained in $U$ and take their union. Since they’re all contained in $U$ the union will be as well. And since arbitrary unions of open sets are still open, it’s an open set. In fact, it’s the largest open set contained in $U$, because it contains all the other open subsets of $U$. We call this the “interior” of $U$, denoted $\mathrm{int}(U)$ or $U^\circ$. Clearly the interior of an open set it the set itself.

Dually, we know that there is at least one closed set containing $U$: the whole space $X$. Then the intersection of all the closed sets containing $U$ will be a closed set containing $U$, and will be the smallest such closed set. We call this the “closure” of $U$ and write $\mathrm{Cl}(U)$ or $\overline{U}$. As for the interior, the closure of a closed set is the set itself.

Now the complement of the closure of $U$ is an open set contained in the complement of $U$. In fact, any other open set contained in the complement of $U$ will be contained in this one, so it is the interior of the complement of $U$. Dually, the closure of the complement of $U$ is the complement of the interior of $U$.

We can write this fact down categorically as well. Since it reverses subset containment, complementation is a contravariant equivalence from the poset $P(X)$ of subsets of $X$ (considered as a category) to itself. That is, $P(X)$ is equivalent to $P(X)^\mathrm{op}$. The interior and closure operators are covariant functors from $P(X)$ to itself, since they preserve containment. The previous paragraph states that these two functors are dual to each other, in the sense that $\mathrm{Cl}^\mathrm{op}:P(X)^\mathrm{op}\rightarrow P(X)^\mathrm{op}$ is the same functor as $\mathrm{int}:P(X)\rightarrow P(X)$ under the above equivalence. So all the really important information is contained in the closure functor.

Now, what do we know about this functor? Well, since $U$ is contained in $\mathrm{Cl}(U)$ we have a natural transformation $\eta:1_{P(X)}\rightarrow\mathrm{Cl}$. Then since $\mathrm{Cl}(\mathrm{Cl}(U))$ is contained in $\mathrm{Cl}(U)$ we have a natural transformation $\mu:\mathrm{Cl}^2\rightarrow\mathrm{Cl}$. I haven’t really covered these yet, but it’s straightforward from here to verify that $\mathrm{Cl}$ along with these two natural transformations forms a monad. If you’re interested in learning more right away, go check out The Catsters’ series of YouTube videos.

We also can easily check that $\mathrm{Cl}(U\cup V)=\mathrm{Cl}(U)\cup\mathrm{Cl}(V)$, and that $\mathrm{Cl}(\varnothing)=\varnothing$. That is, the functor $\mathrm{Cl}$ preserves all finite coproducts. It turns out that this is enough to characterize the topology in its entirety!

Given a set $X$, a closure operator on $X$ is a monad $(\mathrm{Cl},\eta,\mu)$, where $\mathrm{Cl}:P(X)\rightarrow P(X)$ is a functor which preserves finite coproducts. This data is equivalent to the four axioms given by Kuratowski:

1. $U\subseteq\mathrm{Cl}(U)$
2. $\mathrm{Cl}(\mathrm{Cl}(U))=\mathrm{Cl}(U)$
3. $\mathrm{Cl}(U\cup V)=\mathrm{Cl}(U)\cup\mathrm{Cl}(V)$
4. $\mathrm{Cl}(\varnothing)=\varnothing$

From here we can define the closed sets of $X$ to be those in the image of the functor $\mathrm{Cl}$. From axiom 1 we see that $X\subseteq\mathrm{Cl}(X)$, but this closure must be a subset of $X$, and so $X$ is closed. Axiom 4 tells us straight off that $\varnothing$ is closed. Axiom 3 tells us that the finite union of closed sets is closed. We just need to know that arbitrary intersections of closed sets are again closed.

For this, we note that given any collection $\{U_\alpha\}$ of subsets, the intersection $\bigcap\limits_\alpha U_\alpha$ lies in each $U_\alpha$, and so by functoriality $\mathrm{Cl}(\bigcap\limits_\alpha U_\alpha)\subseteq\mathrm{Cl}(U_\alpha)$ for each $\alpha$. Thus we see that $\mathrm{Cl}(\bigcap\limits_\alpha U_\alpha)\subseteq\bigcap\limits_\alpha\mathrm{Cl}(U_\alpha)$. In particular, if all the $U_\alpha$ are closed, then $\mathrm{Cl}(\bigcap\limits_\alpha U_\alpha)\subseteq\bigcap\limits_\alpha U_\alpha$. But since $U\subseteq\mathrm{Cl}(U)$ for all $U$, this inclusion is actually an equality, and thus the intersection of the $U_\alpha$ is in the image of the closure functor. And thus we really have constructed the closed sets of a topology on $X$.

November 13, 2007

## Continuous Maps

Okay, we know what a topological space is. As we might expect, these will be the objects of some category $\mathbf{Top}$. So we need morphisms to connect them, and for this we will use the concept of a continuous map. In some sense these will be “closeness preserving”, but as we’ll see they work a little differently than the various algebraic categories we’re used to.

Here’s the definition: a continuous map from the topological space $(X,\tau_X)$ to the topological space $(Y,\tau_Y)$ is a function $f:X\rightarrow Y$ between the underlying sets so that the preimage of an open set is open. That is, for every open set $U\in\tau_Y$ we can construct $f^{-1}(U)=\{x\in X|f(x)\in U\}$, and we require that this set be open in $X$. That is: $f^{-1}(U)\in\tau_X$. We have a special term for isomorphisms in this category: “homeomorphism”. That is, a homeomorphism between topological spaces is a continuous map with a continuous inverse.

This always feels a little weird to me. Over in algebra we take sets, add extra structure, and define the morphisms to be those functions which preserve that structure. Here we’re defining the morphisms to be those which reflect the extra structure. But let’s accept it and move on.

Of course we also want to consider the categorical perspective. What does a continuous map give us in terms of the topology? It’s a functor from the topology on $Y$ to the topology on $X$, considered as categories! Indeed, if $U\subseteq V$ are two open sets in $Y$, then $f^{-1}(U)\subseteq f^{-1}(V)$ — everything that lands in $U$ also lands in $V$ — and so we send open sets to open sets and inclusion arrows to inclusion arrows. This shows that $f^{-1}:\tau_Y\rightarrow\tau_X$ is a functor.

Even more is true. First, notice that everything in $X$ goes to some point of $Y$, so $f^{-1}(Y)=X$. Similarly, nothing lands in the empty subset of $Y$, so $f^{-1}(\varnothing)=\varnothing$. Given a family of open sets $U_\alpha\in\tau_Y$, a point that lands in their union is in the preimage of at least one of the sets — $f^{-1}(\bigcup\limits_\alpha U_\alpha)=\bigcup\limits_\alpha f^{-1}(U_\alpha)$. Similarly we see that given a finite collection of open sets $U_i\in\tau_Y$, a point that lands in their intersection is in the preimage of each of them — $f^{-1}(\bigcap\limits_{i=1}^nU_i)=\bigcap\limits_{i=1}^nf^{-1}(U_i)$. That is, $f^{-1}$ preserves the finite products and arbitrary coproducts that we assume to exist in these categories.

The catch here is that, as far as we’ve come, we can’t really recover a function $f$ from this functor $f^{-1}$. That is, $f^{-1}$ only talks about open sets rather than about points of our spaces. As yet, these two definitions aren’t quite equivalent, and we’re stuck with talking about the map $f$ as fundamental and deriving from it the functor $f^{-1}$.

November 12, 2007

## Topologies as Categories

Okay, so we’ve defined a topology on a set $X$. But we also love categories, so we want to see this in terms of categories. And, indeed, every topology is a category!

First, remember that the collection of subsets of $X$, like the collection of subobjects on an object in any category, is partially ordered by inclusion. And since every partially ordered set is a category, so is the collection of subsets of $X$.

In fact, it’s a lattice, since we can use union and intersection as our join and meet, respectively. When we say that a poset has pairwise least upper bounds it’s the same as saying when we consider it as a category it has finite coproducts, and similarly pairwise greatest lower bounds are the same as finite products. But here we can actually take the union or intersection of any collection of subsets and get a subset, so we have all products and coproducts. In the language of posets, we have a “complete lattice”.

So now we want to talk about topologies. A topology is just a collection of the subsets that’s closed under finite intersections and arbitrary unions. We can use the same order (inclusion of subsets) to make a topology into a partially-ordered set. In the language of posets, the requirements are that we have a sublattice (finite meets and joins, along with the same top and bottom element) with arbitrary meets — the topology contains the least upper bound of any collection of its elements.

And now we translate the partial order language into category theory. A topology is a subcategory of the category of subsets of $X$ with finite products and all coproducts. That is, we have an arrow from the object $U$ to the object $V$ if and only if $U\subseteq V$ as subsets of $X$. Given any finite collection $\{U_i\}_{i=1}^n$ of objects we have their product $\bigcap\limits_{i=1}^nU_i$, and given any collection $\{U_\alpha\}_{\alpha\in A}$ of objects we have their coproduct $\bigcup\limits_{\alpha\in A}U_\alpha$. In particular we have the empty product — the terminal object $X$ — and we have the empty coproduct — the initial object $\varnothing$. And all the arrows in our category just tell us how various open sets sit inside other open sets. Neat!

November 9, 2007

## (Not) The Tensorator for Span 2-categories

Part of the disappointment I mentioned is that the road I was on just looked so pretty. I’ve said in various places that I agree with (what I understand to be) David Corfield’s view of mathematics as a process of telling good stories, and this was a great story, but unfortunately it just doesn’t quite ring true. Before I purge it, I want to show you the picture of the tensorator as I thought it would work.

Across the top are two tensor products of one span and one object each, and across the bottom are the other two, giving the compositions in both orders. The squares (that look like triangles) at the top and bottom are pullbacks, giving the actual composite spans. Then we can put the tensor product $F\otimes G$ in the middle, and get arrows up and down from the universal properties of the pullback squares. And it even looks like a big tensor product symbol!

But ultimately there’s no way to make this span we get always be unitary, or even invertible. And all the pretty pictures in the world can’t save a deeply flawed story. Just ask Michael Bay.

## The Tensorator for Span 2-categories

I’ve just had a breakthrough today on my project to add structures to 2-categories of spans. I was hoping to generalize from the case of a monoidal structure on the base category $\mathcal{C}$ that preserved pullbacks. After some discussions with John Baez and Todd Trimble (to whom I’m much indebted), I set off on this new quest and ran into some difficulties. Finally I’ve established that in order to have a well-behaved tensorator we must assume that the monoidal structure preserves pullbacks! This is a bit of a downer, in that I was really hoping to construct a wider class of braided monoidal 2-categories with duals, but at least it covers the cases that originally drew me to the problem. With luck there will still be something interesting here. Anyhow, let’s see how this works.

First we have to consider a span $B\xleftarrow{f}A\xrightarrow{g}C$. We want this to be invertible, and further we want its inverse to be its reflection $C\xleftarrow{g}A\xrightarrow{f}B$. When we pull back $g$ over itself we get a square

Here we can swap $x_l$ and $x_r$ to get a unique arrow $\beta:B\rightarrow B$ with $x_l\circ\beta=x_r$. For this to give us the identity span on $B$ we need to have $f\circ x_l=1_B=f\circ x_l\circ\beta$. This tells us that $\beta=1_B$, and we will say $f'$ for the equal arrows $x_l=x_r$, which is a right inverse for $f$: $f\circ f'=1_B$. Similarly we find a right inverse $g'$ for $g$.

Now we use the universality of the pullback in the diagram

to give us a unique arrow $x$ with $f'\circ x=g'$, and similarly a unique arrow $y$ with $g'\circ y=f'$. Then we see that $f'\circ 1_B=f'=f'\circ x\circ y$, and the universality condition tells us that $1_B=x\circ y$, while $1_C=y\circ x$. And thus for the span we started with to have an inverse of the right form we must have $B\cong C$.

Now we look for a tensorator $\bigotimes_{f,g}:(f\otimes B')\circ(A\otimes g)\Rightarrow(A'\otimes g)\circ(f\otimes B)$. We start with spans
$f=(A\xleftarrow{f_l}F\xrightarrow{f_r}A')$
$g=(B\xleftarrow{g_l}G\xrightarrow{g_r}B')$
and we must find a span between the pullback objects $(F\otimes B')\circ(A\otimes G)$ and $(A'\otimes G)\circ(F\otimes B)$. Further, we will want this span to have its own reflection for an inverse, as above. But as we just showed, this means that the two objects at the ends of its legs must always be isomorphic.

Now we can specialize to pick $A'=F$, $f_r=1_F$, $B=G$, and $g_l=1_G$. Then the one leg of the tensorator span will be $F\otimes G$, and so the other leg must be as well, no matter what we choose for $f_l$ and $g_r$! That is, to have any hope of finding such a well-behaved tensorator, the monoidal product on $\mathcal{C}$ must preserve pullbacks!

November 7, 2007 Posted by | Category theory | 1 Comment

## Topology

Well, I’m not quite done with the updates of the topics, but I’ve gotten a number of other things done on my break. Now there’s a search bar over on the right, and the WordPress bug for subtopics has been handled. Rather than delay any longer, I guess I should jump back into the thick of it.

Topology is, roughly speaking, the study of spaces where we have an idea of what it means for points to be “close” to each other, and functions which “preserve closeness”. We don’t care about anything but the most general notion of shape. There’s the famous example of a coffee mug and a doughnut being “the same” to a topologist because they both have one hole, and if you make them out of clay you can deform one into the other without making any drastic changes like a sharp cut. In fact, it’s common to say that topology is all about situations like this, where our shapes are made from clay or rubber sheets that can be deformed around, but as we’ll see there are plenty of situations where we can make cuts (as long as we sew them up again nicely) or even weirder things can happen. Deformations are a good intuition for some aspects of topology, but they’re definitely not the most general.

Okay, so how can we get a handle on this notion of “closeness”. The usual way is to take the set of points $X$ we’re looking at and define some collection $T\subseteq P(X)$ of its subsets as the “open” subsets. Such a collection is required to satisfy a few rules:

• The empty set $\varnothing$ and the whole set $X$ are both in $T$
• The union $\bigcup\limits_\alpha U_\alpha$ of any collection $\{U_\alpha\}\subseteq T$ of subsets in $T$ is again in $T$
• The intersection $\bigcap\limits_{i=1}^nU_i$ of any finite collection $\{U_i\}\subseteq T$ of subsets in $T$ is again in $T$

We call the specified collection $T$ a “topology” on the set $X$, and pair of a set $X$ and a topology $T$ on $X$ we call a “topological space. The elements of $T$ we call the open sets of $X$, and their complements in $X$ we call the closed sets.

Notice here that the collection of closed sets is completely determined by the collection of open sets. This leads to an alternate viewpoint, where we define a collection $T\subseteq P(X)$ of subsets of $X$ satisfying:

• The empty set $\varnothing$ and the whole set $X$ are both in $T$
• The intersection $\bigcap\limits_\alpha U_\alpha$ of any collection $\{U_\alpha\}\subseteq T$ of subsets in $T$ is again in $T$
• The union $\bigcup\limits_{i=1}^nU_i$ of any finite collection $\{U_i\}\subseteq T$ of subsets in $T$ is again in $T$

Now the elements of $T$ are called the closed subsets of the topological space, and their complements are called the open subsets.

We can put more than one topology on the same set $X$, and we can compare different topologies. Let’s say that we have topologies $T_1$ and $T_2$ on a set $X$, so that $T_1\subseteq T_2$. That is, every subset of $X$ that $T_1$ calls open, $T_2$ does as well. In this case, we say that the topology $T_1$ is “coarser” than $T_2$, or that $T_2$ is “finer” than $T_1$. Since we define this relationship by restricting subset containment from $P(P(X))$ to those collections of subsets of $X$ which are actually topologies, it defines a partial order on the collection $\mathcal{T}$ of all topologies on $X$.

The coarsest possible topology is $\{\varnothing,X\}$, which says that only the empty subset and the whole set are open. We call this the “trivial” or the “indiscrete” topology on $X$. Conversely, the finest possible topology is $P(X)$, which says that every subset is open. This we call the “discrete” topology on $X$. Useful topologies tend to fall somewhere between these two extremes, but at least we know that $\mathcal{T}$ has a top and a bottom element for the coarseness relation.

In the middle, let’s say we have some collection $\{T_\alpha\}\subset\mathcal{T}$ of topologies. Then we can define their intersection $\bigcap\limits_\alpha T_\alpha$ as subsets of $P(X)$. This will also be a topology, as is easily shown from the definition above. It is the finest topology which is coarser than all the topologies in $\{T_\alpha\}$, and so any subset of $\mathcal{T}$ has a greatest lower bound.

On the other hand, the union of this collection may not be a topology, which could serve as a least upper bound. However, there is always at least one topology that contains this union — the discrete topology. So we can consider the collection — known to be nonempty — of all topologies which contain the union $\bigcup\limits_\alpha T_\alpha$. The intersection of this collection of topologies will be a topology (as above) which is finer than each topology $T_\alpha$, and is the coarsest possible such topology. Thus any subset of $\mathcal{T}$ has a least upper bound.

Together, these results say that the $\mathcal{T}$ is a complete lattice under the coarseness relation. This turns out to be useful when we have some set we want to put a topology on, and we want to do it in the coarsest possible way subject to a collection of requirements. The fact that $\mathcal{T}$ is a complete lattice says that we can find the coarsest possible topology satisfying the relations one at a time, and then we can find the coarsest topology finer than each of them.

November 5, 2007