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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?

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  • $\begingroup$ Sorry. Is some part of the question unclear? $\endgroup$
    – L. R.
    Apr 15 at 10:44
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    $\begingroup$ 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. $\endgroup$ Apr 15 at 11:37

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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.

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