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I am aware of this question and this one, the answers to which show that the natural numbers can be equipped with a compact Hausdorff topology. But what happens if one also requires the operation $$ \max: {\Bbb N} \times {\Bbb N} \to {\Bbb N} $$ to be continuous?

My intuition is that it is not possible because the profinite approach fails. More precisely, for each natural $t$, let ${\Bbb N}_t = \{0, 1, \ldots, t \}$. Then $({\Bbb N}_t, \max)$ is a finite (and hence compact Hausdorff for the discrete topology) monoid. Furthermore, the map $$ f_t: ({\Bbb N}, \max) \to ({\Bbb N}_t, \max) \text{ defined by } f_t(x) = \min(x, t) $$ is a monoid morphism. Thus one can embed $({\Bbb N}, \max)$ into the compact Hausdorff product monoid $\prod_{t \in {\Bbb N}}({\Bbb N}_t, \max)$ by identifying $n$ with $(0, 1, 2, \ldots, n, n, n, \ldots)$. Unfortunately, $({\Bbb N}, \max)$ is not closed for this topology, since the sequence $u_n = (0, 1, 2, \ldots, n, n, n, \ldots)$ converges to $(0, 1, 2, \ldots, n, n+1, n+2, \ldots)$.

But perhaps some other topology would work... I also tried to use the fact that a countable compact Hausdorff space contains at least one isolated point, but it does not seem to help. I am probably missing some elementary argument, and help would be welcome.

J.-E. Pin
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1 Answers1

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Suppose for contradiction that we have found a topology on $\mathbb{N}$ making it a compact Hausdorff monoid wrt $\max$.

Since $\mathbb{N}$ is infinite and compact, it cannot be the case that every singleton subset of $\mathbb{N}$ is open. So, let $m$ be a natural number such that $\{m\}$ is not open.

Also, we can write $\mathbb{N}$ as the union $\bigcup_{n \in \mathbb{N}} \{n\}$ of closed sets, and thus (by the Baire category theorem) there is some $n_0 \in \mathbb{N}$ such that $\{n_0\}$ is open. Then $\mathbb{N} \setminus \{n_0\}$ is again compact Hausdorff, and we can repeat the argument to obtain another point $n_1$ such that $\{n_1\}$ is open. Etc., so we conclude that there are infinitely many clopen points in $\mathbb{N}$.

So, let $N \in \mathbb{N}$ be such that $m < N$ and $\{N\}$ is open. Then $\max^{-1}(\{N\})$ is open in $\mathbb{N} \times \mathbb{N}$. Since $(m,N) \in \max^{-1}(\{N\})$, there exist open sets $U,V \subseteq \mathbb{N}$ such that $(m,N) \in U \times V \subseteq \max^{-1}(\{N\})$. In particular, $U$ and $V$ are finite.

Since $U$ is finite and open, we have that $\{u\}$ is open for all $u \in U$ (using the fact that $\mathbb{N}$ is Hausdorff). But also we have $m \in U$, so we conclude that $\{m\}$ is open, contradicting our choice of $m$. $\square$

diracdeltafunk
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