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If we let $\mathbb{N}^*$ be the set of ultrafilters on $\mathbb{N}$, $F_n = \{ A \subset \mathbb{N} : n \in A \}$ for each $n \in \mathbb{N}$, and $N = \{ F_n : n \in \mathbb{N} \}$, how can I show that if $f: N \rightarrow Y$ is any function from $N$ into a compact Hausdorff space $Y$, then there exists a continuous function $f^{*}: \mathbb{N}^* \rightarrow Y$ such that $f^{*} |_{N} = f$.

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up vote 3 down vote accepted

Let $p\in\mathbb{N}^*$, and let $\sigma=\langle y_n:n\in\mathbb{N}\rangle$ be a sequence in $Y$. Say that $y\in Y$ is the $p$-limit of $\sigma$, written $$y=p\text{ -}\lim_n y_n\;,$$ if $\{n\in\mathbb{N}:y_n\in U\}\in p$ for every nbhd $U$ of $y$. If you think of the members of $p$ as ‘large’ subsets of $\mathbb{N}$, this says that $y$ is the $p$-limit of $\sigma$ iff every nbhd of $y$ contains a ‘large’ set of the terms of $\sigma$. Since $Y$ is compact and Hausdorff, every sequence in $Y$ has a unique $p$-limit.

Uniqueness is an easy consequence of the fact that $Y$ is Hausdorff. To show existence, let $\sigma=\langle y_n:n\in\mathbb{N}\rangle$ be a sequence in $Y$, and let $p\in\mathbb{N}^*$. For each $U\in p$ let $$A_U=\{n\in\mathbb{N}:y_n\in U\}\;,$$ and let $K_U=\operatorname{cl}A_U$; the family $\mathscr{K}\triangleq\{K_U:U\in p\}$ is centred (i.e., has the finite intersection property), and $Y$ is compact, so $\bigcap\mathscr{K}\ne\varnothing$. Suppose that $y\in\bigcap\mathscr{K}$, and let $V$ be a nbhd of $y$. Let $B_V=\{n\in\mathbb{N}:y_n\in V\}$; then $B_V\cap A_U\ne\varnothing$ for each $U\in p$, and $p$ is an ultrafilter, to $B_V\in p$, and $y$ is therefore a $p$-limit of $\sigma$.

Now let $f:N\to Y$ be any function. Let $\sigma=\langle f(F_n):n\in\mathbb{N}\rangle$. We just saw that for each $p\in\mathbb{N}^*$ the sequence $\sigma$ has a unique $p$-limit. Extend $f$ to $f^*$ by setting $$f^*(F_n)=f(F_n)$$ for $n\in\mathbb{N}$ and $$f^*(p)=p\text{ -}\lim_n f(F_n)$$ for $p\in\mathbb{N}^*$. You shouldn’t have too much trouble showing that this $f^*$ is continuous.

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Thank you @Brian, I did mean into, not onto. I made the correction. – josh Feb 19 '12 at 22:46

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