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The question is exactly the title. Is there a good classification of which functions from $\mathbb{N}$ to $\mathbb{N}$ (or, more generally, from $\mathbb{N}^n$ to $\mathbb{N}$)? Also, what is a good source to learn about $\beta\mathbb{N}$?

I think I can prove that every function extends uniquely, but this seems a little strong, especially since I can't find this fact anywhere.

(My argument is roughly this: since $\mathbb{N}$ is dense in $\beta\mathbb{N}$, we only need to show extension; uniqueness is immediate. Let $f: \mathbb{N}\rightarrow\mathbb{N}$. Define a new function $\hat{f}: \beta\mathbb{N}\rightarrow\beta\mathbb{N}$ by $\hat{f}(\mathcal{U})=\lbrace X: \exists A\in \mathcal{U}(f(A)\subseteq X)\rbrace$. This extends $f$, and takes values in $\beta\mathbb{N}$, so the only thing to check is that it is continuous. To see this, take some $B\subseteq \mathbb{N}$; we need to show that the $\hat{f}$-preimage of $\lbrace \mathcal{U}: B\in\mathcal{U}\rbrace$ is open. But the preimage of a single ultrafilter is open, so this is clear. As a subquestion, is this argument correct? Or salvagable?)

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By general abstract nonsense, a continuous map $\beta \mathbb{N} \to \beta \mathbb{N}$ is the same thing as a map-of-sets $\mathbb{N} \to \beta \mathbb{N}$. (Use the universal property of Stone–Čech compactification.) –  Zhen Lin Aug 31 '12 at 5:47
    
Usually when people talk about extending continuous functions they mean enlarging the domain without changing the codomain, but you seem to want to change the codomain as well. –  Qiaochu Yuan Aug 31 '12 at 5:55
    
I've added compactness tag, since the question deals with Stone-Cech compactification. I was hesitating, whether to add filters tag - $\beta\mathbb N$ can be viewed as a set of ultrafilters and this is obviously the representation which the OP is using. (Although there are many various ways Stone-Cech compactification can be defined.) I've decided not to add this tag and make a comment - thus leaving the decision to the OP. –  Martin Sleziak Aug 31 '12 at 6:51

2 Answers 2

You were doing fine until you got to the proof of continuity. For that you want the following calculation, where for $A\subseteq\Bbb N$ I set $\widehat A=\{\mathscr{U}\in\beta\Bbb N:A\in\mathscr{U}\}$.

$$\begin{align*} \hat f^{-1}\left[\widehat B\right]&=\left\{\mathscr{U}\in\beta\Bbb N:B\in\hat f(\mathscr{U})\right\}\\ &=\left\{\mathscr{U}\in\beta\Bbb N:\exists U\in\mathscr{U}\Big(f[U]\subseteq B\Big)\right\}\\ &=\left\{\mathscr{U}\in\beta\Bbb N:\exists U\in\mathscr{U}\Big(U\subseteq f^{-1}[B]\Big)\right\}\\ &=\left\{\mathscr{U}\in\beta\Bbb N:f^{-1}[B]\in\mathscr{U}\right\}\\ &=\widehat{f^{-1}[B]} \end{align*}$$

Suppose that $f:\Bbb N\to X$ is a sequence in some compact Hausdorff space $X$, and $\mathscr{U}\in\beta\Bbb N$. For each $U\in\mathscr{U}$ let $K_U=\operatorname{cl}_Xf[U]$, and let $\mathscr{K}=\{K_U:U\in\mathscr{U}\}$; clearly $\mathscr{K}$ is centred, so $\bigcap\mathscr{K}\ne\varnothing$. Let $x\in\bigcap\mathscr{K}$, and let $V$ be an open nbhd of $x$. Let $A=\{n\in\Bbb N:f(n)\in V\}$. If $A\notin\mathscr{U}$, let $U=\Bbb N\setminus A\in\mathscr{U}$; then $V\cap K_U=\varnothing$ a contradiction. Thus, $\{x\in\Bbb N:f(n)\in V\}\in\mathscr{U}$ for each open nbhd $V$ of $x$. Since $X$ is Hausdorff, this easily implies that $\bigcap\mathscr{K}=\{x\}$, and we write $x=\mathscr{U}\text{-}\lim f$.

What you’re doing is letting $\hat f(\mathscr{U})=\mathscr{U}\text{-}\lim f$ for each $f:\Bbb N\to\Bbb N$, taking $X$ to be $\beta\Bbb N$. This construction is one way to prove the general fact that every function $f:\Bbb N$ to a compact Hausdorff space extends to $\beta\Bbb N$. Of course if you know that the Čech-Stone compactification $\beta X$ has (and is characterized among compactifications by) the property that any continuous function from $X$ to a compact Hausdorff space extends to $\beta X$, then you don’t have to go through the explicit calculations. It sounds from your question, though, as if you’re also interested in the nuts and bolts of $\beta\Bbb N$.

Added: What would be a good introduction to $\beta\Bbb N$ depends a lot on your interests. Are you interested in it from a primarily set-theoretic point of view, with particular interest in different types of ultrafilters, or is your interest more topological, so that you’re interested in Čech-Stone compactifications generally? For the latter there’s always the classic Rings of Continuous Functions, by Gillman & Jerison; it’s dated, but there’s still a lot of good basic material there. A little less dated is The Theory of Ultrafilters, by Comfort & Negrepontis. Jan van Mill’s chapter, An introduction to $\beta\omega$, in the Handbook of Set-Theoretic Topology, ed. K. Kunen & J.E. Vaughan, requires a bit of background but is well worth reading once you have the background. My library is a bit old now, and there may well be some good treatments that are more recent than any with which I’m familiar.

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In connection with Ramsey theorem, Andres Caicendo recommended N. Hindman and D. Strauss, Algebra in the Stone-Cech compactification here. From the parts of this book I have read so far, I'd say it might be useful for someone who wants to apply $\beta\omega$ in proofs of combinatorial results. –  Martin Sleziak Aug 31 '12 at 6:43
    
@Martin: Thanks. I meant to mention that one in connection with combinatorics and then forgot. –  Brian M. Scott Aug 31 '12 at 6:45

There's a more general principle at work here. Taking Stone–Čech compactifications is functorial: it not only assigns to every topological space $X$ a compact Hausdorff space $\beta X$, it assigns to every continuous map of topological spaces $f : X \to Y$ a continuous map $\beta f : \beta X \to \beta Y$, and it does so in a way that is compatible with composition.

This can be seen in several way depending on exactly how you construct the Stone–Čech compactification. Morally the reason the Stone–Čech compactification is functorial is that it is left adjoint to the inclusion functor from compact Hausdorff spaces to spaces, although proving that such a left adjoint exists requires some work.

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