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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?

t.b.
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skf23852
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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 Answers2

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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).

Asaf Karagila
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  • By $f''(U)$, do you mean $f(U)$? – Zev Chonoles Aug 31 '11 at 17:11
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    @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
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    @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
  • Thanks. It seems to me that $\operatorname{Rng}(f|_U)$ is about as complicated as what it is supposed to abbreviate... – t.b. Aug 31 '11 at 17:25
  • @Theo: What do you think it is supposed to abbreviate? I always though that $f''U$ abbreviates $\operatorname{Rng}(f|_U)$ to begin with :-) – Asaf Karagila Aug 31 '11 at 19:02
  • Not the other way around? :) I count 10 symbols in ${f(x):x \in U}$ while $\operatorname{Rng}(f|_U)$ contains 8. Not to mention the texed version... – t.b. Aug 31 '11 at 19:05
  • @Theo: Oh, yeah. If you think about it this way. I always thought about the former as the definition of the latter, and $f''U$ as the abbreviation thereof :-) – Asaf Karagila Aug 31 '11 at 19:11
  • You know, I'm still too much of an analyst that I could ever consider abusing primes this way... Anyway, I guess de gustibus... – t.b. Aug 31 '11 at 19:22
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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 $\mathbb N$.

Greg Graviton
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