# The Unapologetic Mathematician

## Continuity redux

So now we have two new ways to talk about topologies: neighborhoods, and closure operators. We can turn around and talk about continuity directly in our new languages, rather than translating them into the open set definition we started with.

First let’s tackle neighborhoods. Remember that a continuous function $f$ from a topological space $(X,\tau_X)$ to $(Y,\tau_Y)$ is one which pulls back open sets. That is, to every open set $V\in\tau_Y$ there is an open set $f^{-1}(V)\in\tau_X$ which $f$ sends into $V$. But in the neighborhood definition we don’t have open sets at the beginning; we just have neighborhoods of points.

What we do is notice that a neighborhood $N$ of a point $y$ is a set which contains an open set $V$ containing $x$. In particular we can consider neighborhoods of a point $f(x)$. The the preimage $f^{-1}(V)$ is an open set containing $x$, which is a neighborhood! So, given a neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ so that $f(U)\subseteq V$. This is an implication of the definition of continuity written in the language of neighborhoods, and it turns out that we can turn around and derive our definition of continuity from this condition.

To this end, we consider sets $X$ and $Y$ with neighborhood systems $\mathcal{N}_X$ and $\mathcal{N}_Y$, respectively. We will say that a function $f:X\rightarrow Y$ is continuous at $x$ if for every neighborhood $V\in\mathcal{N}_Y(f(x))$ there is a neighborhood $U\in\mathcal{N}_X(x)$ so that $f(U)\subseteq V$, and that $f$ is continuous if it is continuous at each point in $X$.

Now, let $V$ be an open set in $Y$. That is, a set which is a neighborhood of each of its points. We must now show that $f^{-1}(V)$ is a neighborhood of each of its points. So consider such a point $x\in f^{-1}(V)$, and its image $f(x)\in V$ Since we are assuming that $V$ is a neighborhood of $f(x)$, there must be a neighborhood $U$ of $x$ so that $f(U)\subseteq V$. But then $U\subseteq f^{-1}(V)$, and since the neighborhoods of $x$ form a filter this means $f^{-1}(V)$ is a neighborhood as well. Thus the preimage of an open set is open.

In particular, we can consider a set $T\subseteq Y$ and its interior $\mathbf{int}_Y(T)$, which is an open set contained in $T$. And so its preimage $f^{-1}(\mathrm{int}_Y(T))$ is an open set contained in $f^{-1}(T)$. Thus we see that $f^{-1}(\mathrm{int}_Y(T))\subseteq\mathrm{int}_X(f^{-1}(T))$. Finally, we can dualize this property to see that $f(\mathrm{Cl}_X(S))\subseteq\mathrm{Cl}_Y(f(S))$. That is, the image of the closure of $S$ is contained in the closure of the image of $S$ for all subsets $S\subseteq X$. Let’s now take this as our definition of continuity, and derive the original definition from it.

Well, first let’s just dualize this condition to get back to say that $f^{-1}(\mathrm{int}_Y(T))\subseteq\mathrm{int}_X(f^{-1}(T))$ for all sets $T\subseteq Y$. Now any open set $V$ is its own interior, so $f^{-1}(V)\subseteq\mathrm{int}_X(f^{-1}(V))$. But $\mathrm{int}_X(f^{-1}(V))\subseteq f^{-1}(V)$ by the definition of the interior. And so $f^{-1}(V)$ is its own interior, and is thus open.

November 16, 2007 - Posted by | Point-Set Topology, Topology