Troubles to showing that two subsets of $\mathcal{P}(\omega)$ are compact.

Question

I have troubles with a problem and I want get a hint to solve it.

Consider the topological space $2^\omega$ with its usual topology (i,e, the topology generated by the cones). Under characteristic functions, we can identify the space $2^\omega$ with $\mathcal{P}(\omega)$ and give to the last one the "same" topology.

Let $K,K'\subseteq\mathcal{P}(\omega)$ compact sets and consider the sets $K'':=\{G\cap G´:G\in K\wedge G'\in K'\}$ and $\bar{K}:=\{B\in \mathcal{P}(\omega):\exists A\in K(A\subseteq B)\}$. I need to show that $\bar{K}$ and $K''$ are compact sets too.

So, the "narural"way to do it is considering open coverings. For instance, let $\{[s_\alpha]\}_{\alpha\in I}$ where $s_\alpha\in[\omega]^{<\omega}$ a open covering of $\bar{K}$ (Here $[s_\alpha]$ denote the cone generated by $S_\alpha$ since point of view of $\mathcal{P}(\omega)$). So, it is clear that this is a covering of $K$ too and, as $K$ is compact we can take $I_0\in [I]^{<\omega}$ such that $K\subseteq\cup_{\alpha\in I_0}[s_\alpha]$. But is it neccesary that this finite covering is a finite covering of $\bar{K}$ too? The problem is that the sets in $\bar{K}$ does not have neccesary the same "stem" of the sets in $K$.

The same problems appear when we deal with $K''$.

Can someone help me?

Answer 1

There are many ways to show this, but here's how I would do it. Can you show that $K''$ and $\bar{K}$ are the images of compact sets under continuous functions? (There should be an obvious choice for $K''$; for $\bar{K}$ this is a bit trickier.)

More details are hidden below.

$K''$ is the image of the compact set $K\times K'\subseteq\mathcal{P}(\omega)\times\mathcal{P}(\omega)$ under the intersection map $\mathcal{P}(\omega)\times\mathcal{P}(\omega)\to \mathcal{P}(\omega)$ which is easily seen to be continuous.

$\bar{K}$ is the image of the set $S=\{(A,B)\in\mathcal{P}(\omega)\times\mathcal{P}(\omega):A\subseteq B\}\cap K\times\mathcal{P}(\omega)\subset \mathcal{P}(\omega)\times\mathcal{P}(\omega)$ under the second projection map $\mathcal{P}(\omega)\times\mathcal{P}(\omega)\to \mathcal{P}(\omega)$. $S$ is compact since it is closed in $\mathcal{P}(\omega)\times\mathcal{P}(\omega)$.