Timeline for Existence of an arbitrary Small positive continuous real Valued Function
Current License: CC BY-SA 3.0
9 events
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| Jul 10, 2012 at 21:42 | comment | added | Joseph Van Name |
The closest way that I'm aware of to generate a space by ultrafilters is as follows. Let $X$ be a set. For $x\in X$, let $C_{x}$ be a set of ultrafilters where $\\{R\subseteq X|x\in R\\}\in C_{x}$. Suppose $(C_{x})_{x\in X}$ is closed under Fubini sums. Then there's a topology on $X$ such that $\bigcap C_{x}$ is the neighborhood filter at $x$. Furthermore, the spaces where every real-valued function is below some upper semicontinuous function are precisely the spaces generated by such systems $(C_{x})_{x\in X}$ where every $C_{x}$ is finite and every ultrafilter is $\sigma$-complete.
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| Jun 27, 2012 at 16:48 | comment | added | François G. Dorais | Joseph, indeed it seems to me that all such spaces are in some way "generated" by spaces like the one I gave as an example. However, I can't formulate a precise meaning for "generated" in this context (so I can't prove it either). If anyone has any thoughts to make this precise, let me know... | |
| Jun 27, 2012 at 4:23 | comment | added | Joseph Van Name | It should be noted that a slight modification of the above proof shows that whenever a space has this property, then the neighborhood filter around every point is the intersection of finitely many $\sigma$-complete ultrafilters. In particular, if $\mu$ is the first measurable cardinal, then the intersection of less than $\mu$ open sets in open. | |
| Jun 21, 2012 at 6:51 | comment | added | Ali Reza | Hello Dear Dorais. Your Proof is Fantastic. You could relate This Problem with existence of measurable cardinal beautifully. Thank you very much for your nice description. Then with your positive answer we could find that we could not look for this Property in the spaces such as $\mathbb{N}$,$\mathbb{Q}$,$[0,\Omega_1)$ and another well Known examples with nonmeasurable cardinal. | |
| Jun 21, 2012 at 6:44 | vote | accept | Ali Reza | ||
| Jun 20, 2012 at 15:48 | comment | added | Joseph Van Name | The space you constructed above in the second paragraph is the a space $Y$ where $X\subseteq Y\subseteq\upsilon X$ where $|Y\subseteq X|=1$ and where $\upsilon X$ is the Hewitt-realcompactification of $X$. | |
| Jun 20, 2012 at 14:08 | comment | added | François G. Dorais | I had missed the Tychonoff requirement in my first reading of the question. I should point out that the space constructed in the second paragraph is indeed Tychonoff. For the converse implication, it makes no difference whether the space is Tychonoff or not. | |
| Jun 20, 2012 at 0:18 | history | edited | François G. Dorais | CC BY-SA 3.0 |
cleaned up typos and such
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| Jun 19, 2012 at 23:48 | history | answered | François G. Dorais | CC BY-SA 3.0 |