Does there exist a compact Hausdorff topology on the natural numbers? 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?
 A: 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). 
A: 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 $\mathbb N$.
