# Is $\mathbb{N}$ with $\zeta^*$ uniformity totally bounded?

I'm stuck in a proof of the Cech-Stone compactification of the natural numbers. Let $\mathbb{N}$ be the natural numbers with the discrete topology and let $\zeta^*$ be the initial uniformity on it made by all functions $f: \mathbb{N} \rightarrow [0,1]$. Now, it is claimed that with this initial uniformity, $\mathbb{N}$ is totally bounded. Why is this ? I tried to prove it (and I managed to do so in case of one function, by which I mean I managed to find a finite subset $F$ of $\mathbb{N}$ such that for one function $f: \mathbb{N} \rightarrow [0,1]$, and one $n \in \mathbb{N}$, $\mathbb{N} = (f \times f)^{-1}(U^{1/n})(F)$, with $U^{1/n}=\{(x,y)\mid d(x,y) < 1/n\}$ ), but I'm stuck in the case of finite intersections.

I'm also interested to know why this doesn't work in case of $\mathbb{Q}$ with the Euclidean topology and all uniformly continuous functions to $[0,1]$

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You might want to change the title; $\mathbb{N}$ by itself usually implies the discrete uniformity... –  Harry Altman Dec 19 '11 at 11:07
Thanks, will do. –  KevinDL Dec 19 '11 at 11:14

You can see $\zeta^*$ as generated by entourages of the form $$U^f_\varepsilon = \{ (n, m) | \| f(n) - f(m) \|_\infty < \varepsilon \}$$

for all $\varepsilon > 0$, $n \in \mathbb{N}$, and $f : \mathbb{N} \rightarrow [0, 1]^n$.

Using the same argument, you should be able to prove that given $\varepsilon$ and $f : \mathbb{N} \rightarrow [0, 1]^n$, there exists a finite $U^f_\varepsilon$-net in $\mathbb{N}$ (you only need to use the fact that $[0, 1]^n$ itself is totally bounded).

Since now the $U^f_\varepsilon$ constitute a fundamental system of entourages for $\zeta^*$, it follows that $\zeta^*$ is totally bounded.

For your second question, I assume you mean "all continuous functions". Since $\mathbb{Q}$ is a metric space, you get the uniformity induced by the metric, which is of course not totally bounded.

EDIT: Oops, this is not true.

To see that, take a ball $B$ of radius $\varepsilon$ centered in $x_0$, then (using the above notation), $U^f_{1/2}(x_0) \subseteq B$, where $f$ is a function such that $f(x_0) = 1$ and $f \equiv 0$ on the complement of $B$.

EDIT: This proves that the initial uniformity induces the Euclidean topology. The initial uniformity is still totally bounded by the same argument as for $\mathbb{N}$.

Now, coming to your real question, using all uniformly continuous functions to generate a uniformity on $\mathbb{Q}$ still yields a totally bounded uniformity (since it is coarser than the initial uniformity). The point is that this uniformity $\mathcal{U}$ does not coincide with the one induced by the metric (let's call it $\mathcal{E}$), because you are limiting yourself to bounded functions. So all three uniformities are distinct, and have the following inclusion relationships:

$$\mathcal{U} \subset \zeta^*,\quad \mathcal{U} \subset \mathcal{E}$$

even though they all induce the same topology.

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You have functions $f_i:\mathbb{N}\to[0,1]$ for $i=1,\dots,n$ and $\epsilon>0$, and you want a finite $F\subseteq \mathbb{N}$ such that for each $k\in\mathbb{N}$ there is an $m\in F$ with $\max\limits_i\,|f_i(k)-f_k(m)|<\epsilon$.

If you can do this for one function, there’s a small trick that lets you do it for $n$ functions: just define $f:\mathbb{N}\to[0,1]^n:k\mapsto \big \langle f_1(k), \dots, f_n(k)\big\rangle$ and use the metric $$d(\langle x_1,\dots,x_n\rangle,\langle y_1,\dots,y_n\rangle)=\max_{1\le i\le n}|x_i-y_i|$$ on $[0,1]^n$ instead of the metric $d(x,y)=|x-y|$ on $[0,1]$.

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What exactly do you want to prove? I have never come across this problem when considering the Stone-Cech compactification of $\mathbb N$.

Anyhow. Consider finitely many functions $f_1,\dots,f_k:\mathbb N\to[0,1]$. For each $i\in\{1,\dots,k\}$ let $\varepsilon_K\gt 0$ (this replaces your $1/n$). For each $i\in\{1,\dots,k\}$ choose a partition $P_i$ of $[0,1]$ into finitely many sets of diameter $\lt\varepsilon_i/2$.

These finitely many partitions induce partitions $Q_i$ on $\mathbb N$. Each $Q_i$ consists of the $f_i$-preimages of the elements of $P_i$. The $Q_i$ have a common refinement $Q$, which is also a partition of $\mathbb N$ into finitely many pieces. Now each element of $Q$ is a set that is mapped by $f_i$ into a set of diameter $\lt\varepsilon_i/2$.

This shows that $\mathbb N$ is a finite union of "small" sets, where small means small wrt to the $f_i$ and the $\varepsilon_i$. (Sorry, I am not that familiar with the notation concerning uniformities.)

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I think you meant $i\in\{1,\ldots,k\}$ there. –  Asaf Karagila Dec 19 '11 at 11:34
Thanks, exactly what I needed. I needed it to conclude that all ultrafilters are Cauchy filters, for which I needed that $\mathbb{N}$ is totally bounded. Any thoughts why it fails for $\mathbb{Q}$ ? –  KevinDL Dec 19 '11 at 11:43
@Asaf: Yes, indeed. I changed this. –  Stefan Geschke Dec 19 '11 at 13:02