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Let $f:X\times Y\mapsto R$ be jointly continuous, where $X$ is some topological space, $Y$ is some countably compact topological space. My question is whether or not the envelop $\phi(x) := \max_{y\in Y} f(x,y)$ is continuous?

I know the answer is yes, if we strengthen countably compact to compact. And I tried to construct a counterexample with $Y = [0, \aleph_1)$ but failed. Any idea will be appreciated!

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In general the answer is no. For showing this claim I must bring two theorems. You could find these theorems in the page 238 of the text "General topology" written by Ryszard Engelking.

The following theorem is due to Isiwata, Nobel, Hager and comfort:

Theorem1: For the Tychonoff spaces $X , Y$ the following are equivalent:

  • The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.

  • Every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ can be continuously extended over $X \times \beta Y$.

  • For every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ the formula $F(x)=sup_{y\in Y}f(x,y)$ defines a continuous function $F:X\rightarrow \mathbb{R}$.

The following theorem is due to "Tamano" which is essential for our claim:

Theorem2: The cartesian product $X\times Y$ of Tychonoff spaces $X , Y$ is pseudocompact if and only if $X$ and $Y$ are pseudocompact and The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.

But for showing our claim as you probably Know there is a countably compact space $X$ which the product space $X\times X$ is not even Pseudocompact.(You could find such example in the chapter9 of the text Rings of continuous functions written by Gillman and jerison)

Then because $X$ is countably compact it is also pseudocompact. and because $X\times X$ is not pseudocompact (as we mentioned above) each of the statements in Theorem1 fails. Then for example there is a bounded continuous function $f:X\times X \rightarrow \mathbb{R}$ so that the function $F(x)=sup_{y\in Y}f(x,y)$ is not continuous.

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  • $\begingroup$ Thanks AliReza. I just can't help asking you what if $Y$ is sequentially compact? The argument you illustrated does not seem to work anymore (since sequentially compact $\times$ pseudocompact is pseudocompact). Any thought? :-) $\endgroup$ – yaoliang Jul 8 '12 at 3:04

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