# Is every zero-dimensional space with no infinite clopen partition pseudocompact?

For this question, we say that a zero-dimensional space $X$ is $\omega$-pseudocompact if every partition of $X$ into clopen sets is finite. In other words, a zero-dimensional space $X$ is $\omega$-pseudocompact if and only if every continuous mapping $f:X\rightarrow\mathbb{N}$ is bounded. One can easily show that a zero-dimensional space $X$ is $\omega$-pseudocompact if and only if every countable clopen cover of $X$ has a finite subcover. For zero-dimensional spaces, is $\omega$-pseudocompactness equivalent to pseudocompactness?

For strongly zero-dimensional spaces, these properties are equivalent and I give a proof right here. Recall that a completely regular space $X$ is strongly zero-dimensional if the Stone-Cech compactification $\beta X$ is zero-dimensional. It is well known that a space $X$ is strongly zero-dimensional if and only if whenever $Z_{1},Z_{2}$ are disjoint zero-sets, then there is a clopen set containing $Z_{1}$ and disjoint from $Z_{2}$. If $X$ is strongly zero-dimensional and $f:X\rightarrow[0,\infty)$ is an unbounded continuous mapping, then since $f^{-1}[0,n],f^{-1}[n+1,\infty)$ are disjoint zero-sets, there is a clopen set $A_{n}$ containing $f^{-1}[0,n]$ but disjoint from $f^{-1}[n+1,\infty)$. The clopen cover $\{A_{n}|n\in\mathbb{N}\}$ has no finite subcover since $A_{0}\cup...\cup A_{n}$ does not intersect $f^{-1}[n+1,\infty]$ for all $n$. Therefore every strongly zero-dimensional space that is not pseudocompact is not $\omega$-pseudocompact, and clearly every pseudocompact space is $\omega$-pseudocompact.

## 1 Answer

I think the answer is yes. to see the sketch of proof you could suppose on the contrary that, $f$ is not bounded and choose $$x_1, x_2,...,x_m,... \in X$$ and$$r_1,r_2,...r_m,...\in \mathbb{R}^+$$ so that $$r_1< |f(x_1)|< r_2<|f(x_2)|< ... r_m< |f(x_m)|< ...$$ Because $X$ is zero-dimensional and $f$ is continuous, for each $m\in \mathbb{N}$ you could find a clopen subset $V_m \subseteq${$x\in X: r_m< |f(x)|< r_{m+1}$}. You could see that for $i\neq j, V_i\cap V_j=\emptyset$. Moreover $$V_0=X-\cup_{i=1}^{\infty}V_{i}$$ which is equal to $\cup_{i=1}^{\infty}(${$x\in X: |f(x)|< r_{i+1}$}$-V_1\cup V_2 \cup... \cup V_i$$)$, and obviously it is open. It yields that $X$ has an infinite partition and this implies a contradiction. because $X$ is $\omega-$ pseudocompact and has no countable partition of clopen sets.