find convergent subnet

Question

I`m trying to learn how to use nets and subnets to prove some theorems. I came across this problem in Banach space book: let K be a compact subset of a topological space X. suppose that $(x_{\alpha})_{\alpha \in I}$ is a net in X such that whenever U is open set that includes K and $\alpha\in I$ , there is a $\beta_{\alpha,U}$ in I such that $\alpha \preceq \beta_{\alpha,U}$ and $x_{\beta_{\alpha,U}} \in U$. prove that some subnet of $(x_{\alpha})$ converges to an element of K.

I tried to do the same proof as in the theorem with compact space that includes convergent subnet, but I just feel it`s wrong and something should be different. Can anyone please help me with it?

Thanks

Answer 1

Let $\nu=\langle x_\alpha:\alpha\in I\rangle$ be a net in $X$. Recall that $\nu$ is frequently in a set $S$ iff for each $\alpha\in I$ there is a $\beta\in I$ such that $\alpha\preceq\beta$ and $x_\beta\in S$. The key step in the problem is the following lemma.

Lemma. Let $x\in X$. If $\nu$ is frequently in every open nbhd of $x$, then $\nu$ has a subnet converging to $x$.

Proof. Assume that $\nu$ is frequently in every open nbhd of $x$, and let $\mathscr{N}$ be the family of open nbhds of $X$. For $\langle\alpha,U\rangle,\langle\beta,V\rangle\in I\times\mathscr{N}$ write $\langle\alpha,U\rangle\sqsubseteq\langle\beta,V\rangle$ iff $\alpha\preceq\beta$ and $U\supseteq V$; then $\langle I\times\mathscr{N},\sqsubseteq\rangle$ is a directed set. By hypothesis there is a function $\varphi:I\times\mathscr{N}\to I$ such that $$\alpha\preceq\varphi(\langle\alpha,U\rangle)\quad\text{and}\quad x_{\varphi(\langle\alpha,U\rangle)}\in U$$ for each $\langle\alpha,U\rangle\in I\times\mathscr{N}$. For each $\alpha\in I$ we have $\alpha\preceq\varphi(\langle\beta,U\rangle)$ whenever $\langle\alpha,X\rangle\sqsubseteq\langle\beta,U\rangle$, so $\mu=\langle x_{\varphi(\langle\alpha,U\rangle)}:\langle\alpha,U\rangle\in I\times\mathscr{N}\rangle$ is a subnet of $\nu$. (Here I’m using the definition of subnet in Kelley’s General Topology.) Now let $U\in\mathscr{N}$ be arbitrary; there is an $\alpha\in I$ such that $x_\alpha\in U$, and $x_{\varphi(\langle\beta,V\rangle)}\in V\subseteq U$ whenever $\langle\alpha,U\rangle\sqsubseteq\langle\beta,V\rangle$, so $\mu$ converges to $x$. $\dashv$

Now suppose that $K\subseteq X$ is compact, and that if $U$ is any open nbhd of $K$, then $\nu$ is frequently in $U$. Suppose further that no subnet of $\nu$ converges to any point of $K$. Then each $x\in K$ has an open nbhd $U_x$ such that $\nu$ is not frequently in $U_x$. This means that for each $x\in K$ there is an $\alpha(x)\in I$ such that $x_\beta\notin U_x$ whenever $\beta\in I$ and $\alpha(x)\preceq\beta$. $K$ is compact, so there is a finite $F\subseteq K$ such that $K\subseteq\bigcup_{x\in F}U_x$, and $I$ is directed, so there is an $\alpha\in I$ such that $\alpha(x)\preceq\alpha$ for each $x\in F$. But then $\bigcup_{x\in F}U_x$ is an open nbhd of $K$, and $x_\beta\notin U$ whenever $\alpha\preceq\beta$. This contradiction shows that $\nu$ must have a subnet converging to some point of $K$.