I found an interesting exercise but I don't know how to solve it.

Given a topological space $X$, suposse that there exist $\lambda$ a cardinal such that $X$ is homeomorphic to a closed subset of $\mathbb{N}^{\lambda}$ ($X$ can be viewed, in fact, as a closed subset of $\mathbb{N}^{\lambda}$). Is there $\kappa$ a cardinal such that $\beta X$ is homeomorphic to a subspace of $\mathbb{N}^{\kappa}$? Here, the subspace is not necesarilly closed nor open. It's only a subspace of $\mathbb{N}^{\kappa}$.

I really don't know how to solve it. Is it true? My first attempt was take $X\subseteq\mathbb{N}^{\lambda}$ with $X$ closed in $\mathbb{N}^{\lambda}$. By the Tychonoffness of $\mathbb{N}^{\lambda}$ then such space can be viewed as a subspace of $[0,1]^{\alpha}$ for some cardinal $\alpha$, i.e., $\mathbb{N}^{\lambda}\subseteq [0,1]^{\alpha}$. Then $X\subseteq \mathbb{N}^{\lambda}\subseteq [0,1]^{\alpha}$. If we take the closure respect to $Z=[0,1]^{\alpha}$, i.e., $$\text{cl}_{Z}(X)\subseteq\text{cl}_{Z}(\mathbb{N}^{\lambda})\subseteq [0,1]^{\alpha}$$ Can be $\text{cl}_{Z}(X)=\beta X$? From here I don't know how to go because $\text{cl}_{Z}(\mathbb{N}^{\lambda})$, in general, is not of the form $\mathbb{N}^{\kappa}$. Any hint? Thanks for your help.

  • 2
    $\begingroup$ This seems not to bad to sort out when $X$ is strongly zero-dimensional. I don't know the answer in general, since there are $\mathbb{N}$-compact spaces of positive topological dimension (see for instance Mysior). btw, note that $\beta X$ is compact, so if it embeds in $\mathbb{N}^\kappa$, then it does so as a closed subspace. $\endgroup$
    – Tyrone
    Nov 17, 2020 at 13:41

1 Answer 1


Recall that a space is called $\mathbb{N}$-compact, if it can be embedded as a closed subset of $\mathbb{N}^{\lambda}$ for some cardinal $\lambda$. Note that a compact subset of $\mathbb{N}^{\lambda}$ is closed in $\mathbb{N}^{\lambda}$, hence $\mathbb{N}$-compact.
With this notation the question is, when $\beta X$ is $\mathbb{N}$-compact. In fact, Tyrone already almost pointed out the answer, namely:

Let X be completely regular. Then $\beta X$ is $\mathbb{N}$-compact, iff $X$ is strongly $0$-dimensional.

Proof: "$\Rightarrow$": Obviously, $\beta X$ is $0$-dimensional. Hence, $X$ is strongly $0$-dimensional.
"$\Leftarrow$": Since $X$ is strongly $0$-dimensional, $\beta X$ is strongly $0$-dimensional. Since it is compact, in particular, realcompact, it is $\mathbb{N}$-compact. (This is a well-known result proven in 1967 by H. Herrlich. I couldn't find an online version of the proof, but it is cited, for instance, here, and also mentioned in the paper of Mysior mentioned by Tyrone.)

The example given in Mysior is $\mathbb{N}$-compact, but not strongly $0$-dimensional, hence $\beta X$ is not $\mathbb{N}$-compact, thus answering the assumption in the negative.


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