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I was wondering whether there exists a compact Hausdorff topology on $\mathbb N$. The only result I was able to find in this context was that, if a set has a topology that is compact, Hausdorff and has no isolated points then the set is uncountable. But what if isolated points are allowed?

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Identify $\mathbb{N}$ with a convergent sequence and its limit point, say in $\mathbb{R}$. –  t.b. Aug 31 '11 at 14:39

2 Answers 2

up vote 8 down vote accepted

Of course there is.

Example: Consider $f\colon\mathbb N\to\mathbb R$ defined as, $f(0)=0, f(n)=\frac1n$. Now let $\tau$ be the topology defined as:

$U\subseteq\mathbb N$ is open if and only if $f''U$ is open in $\mathbb R$.

This clearly corresponds to the metric $d(x,y)=|f(x)-f(y)|$, so this makes $(\mathbb N,\tau)$ a metrizable space. In fact it is completely metrizable since every point except $0$ is isolated, and $0$ is the only limit point.

This can be easily generalized by taking a countable set of real numbers which is bounded and has only countably many limit points, and any bijection whatsoever between this set and $\mathbb N$.

Of course we cannot drop the limitation that there are only countably many limit points, since a compact metric space is complete and all limit points must be inside it.


If you are familiar with ordinals, then you may want to prove the following:

Theorem: Suppose $\beta$ is a successor ordinal, then in the order topology $\beta$ is Hausdorff and compact.

Corollary: Every countable successor ordinal can be given a compatible metric which is complete, and the result is a compact Polish space (separable, metric and complete).

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By $f''(U)$, do you mean $f(U)$? –  Zev Chonoles Aug 31 '11 at 17:11
    
@Zev: Yes. $f''U=\{f(x)\mid x\in U\}$. It is a common notation (in set theory). –  Asaf Karagila Aug 31 '11 at 17:14
    
Interesting; I will remember it. Thanks! –  Zev Chonoles Aug 31 '11 at 17:17
    
@Zev: I was wondering about that, too. Asaf: Why two primes? What is $f'U$, then? The pre-image of $U$? –  t.b. Aug 31 '11 at 17:17
    
@Theo: I don't recall seeing $f'U$ anywhere in set theory, so I can't really tell you that. Thing is that when the function has an unspecified domain, writing $f(U)$ could be the image of the element $U$ (if the function is defined on both $U$ as an element and as a set), so there are several notations $f''U, f[U], \operatorname{Rng}(f|_U)$, etc). –  Asaf Karagila Aug 31 '11 at 17:19

There exists a bijection between the one-point compactification $\mathbb{N}\cup\lbrace\infty\rbrace$ and $\mathbb N$, for instance by mapping

$$ \infty \mapsto 0 \text{ and } n \mapsto n+1 .$$

Use this map to obtain a compact Hausdorff topology on $\mathbb N$.

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