# When is this closed set compact

Apparently in the polish space $$^\omega\omega$$ a closed $$K\subset\hspace{1mm}^\omega\omega$$ is bounded and therefore compact if it is completely below some $$f\in \hspace{1mm}^\omega\omega$$ as in $$K= \{ g \in K: \forall n\in \omega: g(n)\leq f(n) \}$$. I don't really understand this sort of compactness, since often the $$f$$'s bounding those sets are unbounded themselves. I assume the bounded part stems from the fact that the $$g(n), n\leq m$$ for fixed $$m$$ get actually bounded. How is this proven?

• Sorry. Is some part of the question unclear? Apr 15 at 10:44
• Can you see why, for example, $\prod_{n \in \omega} \{0, \dotsc, n\}$ is compact in the product topology? (Don't overthink it!) In the other direction the argument is indeed just that the co-ordinate projections must be bounded. Apr 15 at 11:37

## 1 Answer

Fix $$f\in\omega^\omega$$ and let $$C=\{g\in\omega^\omega\mid g(n)\leq f(n)\text{ for all }n\}$$. We want to show that $$C$$ is compact (it follows immediately that any closed subspace of $$C$$ is compact as well).

The main point here is that the members of $$C$$ can only attain finitely many values in each coordinate, so that $$C$$ is very narrow in some sense.

The easiest way to see that $$C$$ is compact is to note that $$C=\prod_{n\in\omega}\{0,1,\ldots,f(n)\}$$ is a product of compact sets, hence compact.

One can also show that every sequence in $$C$$ has a convergent subsequence. Indeed let $$(g_i)_{i\in\omega}$$ be a sequence in $$C$$. By the pigeonhole principle there must be some $$k_0\leq f(0)$$ such that $$I_0=\{i\in\omega\mid g_i(0)=k_0\}$$ is infinite. Let $$i_0\in I_0$$ be arbitrary.

Using the pigeonhole principle again we can find $$k_1\leq f(1)$$ such that $$I_1=\{i\in I_0\mid g_i(1)=k_1\}$$ is infinite. Let $$i_1\in I_1$$ be arbitrary.

Iterating this construction we produce a sequence $$i_n$$ for each $$n\in\omega$$ with the property that $$g_{i_{n+1}}$$ and $$g_{i_n}$$ agree on the first $$n$$ integers. It follows that $$g_{i_n}$$ converges pointwise, as desired.