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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?

  • 7
    Identify $\mathbb{N}$ with a convergent sequence and its limit point, say in $\mathbb{R}$.2011-08-31

2 Answers 2

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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).

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    You know, I'm still too much of an analyst that I could ever consider abusing primes this way... Anyway, I guess *de gustibus...*2011-08-31
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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 $\mathbb N$.