Heartbeat

I have not written in a while, but this blog is still alive, so I will quickly prove two of my favorite (among many) results of topology. These stem straight from my preparation for the coming preliminary examinations.

School all of a sudden snatched most of my time.

Let $X$ be a topological space.

Claim: $X$ is Hausdorff if and only if $\Delta:=\left\{\left(x,x\right):x\in X\right\}$ is closed.

Proof: Suppose that $X$ is Hausdorff, and let $\left(x_1,x_2\right)\in X\times X\setminus\Delta$. Then $x_1\neq x_2$, which implies that there exist disjoint neighborhoods $U$ and $V$ of $x_1$ and $x_2$, respectively, which implies that $U\times V\subseteq X\times X\setminus\Delta$, which implies that $X\times X\setminus\Delta$ is open, which implies that $\Delta$ is closed.

Suppose that $\Delta$ is closed, and let $x_1,x_2\in X$ such that $x_1\neq x_2$. Then $\left(x_1,x_2\right)\in X\times X\setminus\Delta$, which is open, which implies that there exists a neighborhood $U\times V$ of $\left(x_1,x_2\right)$ contained in $X\times X\setminus\Delta$, which implies that there exist disjoint neighborhoods $U$ and $V$ of $x_1$ and $x_2$, respectively, which implies that $X$ is Hausdorff. $\blacksquare$

Let $X$ and $Y$ be topological spaces, let $Y$ be Hausdorff, let $A\subseteq X$, let $f,g:\overline A\to Y$ be continuous, and let $f\left(x\right)=g\left(x\right)$ for every $x\in A$.

Claim: $f\left(x\right)=g\left(x\right)$ for every $x\in\overline A$.

Proof: Define $h:\overline A\to Y\times Y$ by $x\mapsto\left(f\left(x\right),g\left(x\right)\right)$. Then $h$ is continuous and $h\left(A\right)\subseteq\Delta$, which implies that $h(\overline A)\subseteq\overline{h\left(A\right)}\subseteq\overline\Delta$. Since $Y$ is Hausdorff, $\Delta$ is closed, which implies that $h(\overline A)\subseteq\Delta$, which implies that $f\left(x\right)=g\left(x\right)$ for every $x\in\overline A$. $\blacksquare$

I like these results because the first characterizes Hausdorff spaces in simple terms, and the second shows that no shenanigans happen at the continuous images in Hausdorff spaces of limit points.

Remark: It is implicit in the first claim that the topology on $X\times X$ is the product one, which is generated by the set of products of open subsets of $X$. Let $\left(x_1,x_2\right)\in X\times X$, and let $N$ be a neighborhood of $\left(x_1,x_2\right)$. Then
$$N=\bigcup_{\alpha\in I}U_\alpha\times V_\alpha,$$
where $U_\alpha$ and $V_\alpha$ are open, and $I$ is some index set. Therefore, there exists a $\beta\in I$ such that $\left(x_1,x_2\right)\in U_\beta\times V_\beta$.

No comments:

Powered by Blogger.