Re: question on compact spaces

Stephen J. Herschkorn wrote:

David C. Ullrich wrote:

On Mon, 25 Jun 2007 21:29:00 -0400, "Stephen J. Herschkorn"
<sjherschko@xxxxxxxxxxxx> wrote:


quasi wrote:



On Mon, 25 Jun 2007 20:13:18 -0400, "Stephen J. Herschkorn"
<sjherschko@xxxxxxxxxxxx> wrote:



Mike wrote:

Does anybody know how to prove or disprove the following proposition:

Proposition: Every quasicompact space is the surjective image under a continuous map of a compact Hausdorff space.

Here by quasicompact I just mean that the space satisfies the finite subcovering property. I.e. quasicompact just means what most books call compact, though Bourbaki includes the Hausdorff property in the word compact.

It seems to me that the proposition is true, at least in ZFC: There is a continuous surjection from 2^X to X, your (quasi)compact space.

That doesn't look right.

What topology are you assuming on 2^X?

I was intentionally coy with details. Take my post as a hint. Do you really want me to spill the beans? (A sincere question.)


I do. How do you show that there is a continuous map from 2^X
(presumably with the product topology) onto X?



Actually, while thinking about the problem later, I realized my intended map was incorrect. I had intended an evaluation map, but then realized that my map was from X^2 to X, not from 2^X.

Do you think this approach can be salvaged? My initial thought was a map from a compact subset of some function space (if not 2^X, maybe C(X, [0,1]) or something else) to X. I don't have a lot of time right now, and it's way too early in the day for me, but can we well-order X somw way and find a closed subset K of 2^X such that the map g: K -> X, f |-> min [f^(-1)](0) is continuous and surjective?


Have to run, but how about an evaluation map from X^X -> X? I think that works, but I very well could be wrong in my haste.

--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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