Let $X$ be a pseudocompact space (i.e. $X$ is Tychonoff space and every continuous function $f :X\to \Bbb R$ is bounded) and let $Y$ be a Tychonoff compact or sequentially compact space.

In each case, how to prove that $X\times Y$ is pseudocompact?

  • $\begingroup$ For a continuous $f:X\times Y\to\Bbb R$, what about taking $ m(x):=\min \{f(x,y):y\in Y\}\ $ and $\ M(x):=\max \{f(x,y):y\in Y\}$? These seem to be continuous, thus are bounded. $\endgroup$
    – Berci
    Oct 23 '13 at 20:24
  • $\begingroup$ When I tried to show that this functions are continuous I failed. $\endgroup$ Oct 24 '13 at 5:21

First, we need the following criteria:

A Tychonoff space $X$ is pseudocompact if and only if every locally finite open covering of $X$ has a finite subcovering.

Proof. Assume that X is not pseudocompact, so there is unbounded continuous $f:X\to\mathbb R$. Then $\{f^{-1}((n-1,n+1)):n\in\mathbb Z\}$ is a locally finite open cover of X with no finite subcover. Conversely, assume that $\mathfrak U$ is a locally finite open cover of X with no finite subcover. Let $\{U_n:n\in\mathbb N\}\subseteq\mathfrak U$ be sequence of distinct non-empty members of $\mathfrak U$. Select $x_n\in U_n$ and define continuous $f_n:X\to [0,n]$ such that $f_n(x_n)=n$ and $f_n(X\setminus U_n) = 0$. Then, since $\mathfrak U$ is locally finite, $f=\max\{f_n:n\in\mathbb N\}$ is continuous real-valued function on $X$, so $X$ is not pseudocompact.

Second, we prove that for pseudocompact $X$ and compact $Y$ the product space $X\times Y$ is pseudocompact.

Proof. Let $\mathfrak U$ be an arbitary locally finite open covering of $X\times Y$. Assume that $\varphi:X\times Y\to X$ and $\psi:X\times Y\to Y$ are canonical projections.
For any $x\in X$ let $\mathfrak B_x = \{U\in\mathfrak U:x\in\varphi(U)\}$. Then $\{\psi(B):B\in\mathfrak B_x\}$ is open cover of compact $Y$ so it has a finite subcover $\{\psi(B):B\in\mathfrak B_x'\}$. Therefore $V_x = \bigcap\{\varphi(B):B\in\mathfrak B_x'\}$ is open neighbourhood of $x$ and $\mathfrak V = \{V_x: x\in X\}$ is open cover of $X$.
We have to show that $\mathfrak V$ is locally finite.
For arbitary $(x,y)\in X\times Y$ let us choose a neighbourhood of $(x,y)$ $W_{x,y}$ that intersects only finitely many of the sets in $\mathfrak U$. Then for all $x\in X$ assume $\mathfrak D_x = \{W_{x,y}:y\in Y\}$. The collection $\{\psi(D):D\in\mathfrak D_x\}$ is open cover of $Y$ thus there is its finite subcover $\{\psi(D):D\in\mathfrak D_x'\}$. $W_x = \bigcap\{\varphi(D):D\in\mathfrak D_x'\}$ is neighbourhood of $x$. From its definition we can see that it intersects only finite number of members of $\mathfrak V$. Hence $\mathfrak V$ is locally finite and since $X$ is pseudocompact it has a finite subcover $\mathfrak V'$.
Now we can construct finite open subcover $\mathfrak U'$ of $\mathfrak U$: let $\mathfrak U' = \bigcup\{\mathfrak B_x':V_x\in\mathfrak V'\}$.

And thirdly, the fact that the product $X\times Y$ of pseudocompact $X$ and sequentially compact $Y$ is pseudocompact follows from the previous case. I've taken the proof from Engelking's book (subch. 3.10).

Proof. Suppose that there exists unbounded continuous $f:X\times Y\to\mathbb R$. Select $(x_n,y_n)\in X\times Y$ such that $|f(x_n,y_n)|\geq n \;\;\forall n\in\mathbb N$. $Y$ is sequentially compact thus the sequence $\{y_n:n\in\mathbb N\}$ has a subsequence $\{y_{n_k}:k\in\mathbb N\}$ which converges to some $z\in Y$.
Let $Z:=\{y_{n_k}:k\in\mathbb N\}\cup\{z\}$. $Z$ is compact so $X\times Z$ is pseudocompact, but this leads to contradiction, since $f\restriction_{X\times Z}$ is continuous and unbounded.
Therefore any continuous real-valued function on $X\times Y$ is bounded.


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