# The Unapologetic Mathematician

## 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