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The question is also posted here.

The paper is Mizokami : On characterizations of spaces with $G_\delta$-diagonals

See its Theorem 1, also you can see the picture . http://picpaste.com/a-eaiF4d3t.bmp.

Theorem 1: A space $X$ has a $G_\delta$-diagonal iff there is an open mapping (single valued) $f$ from a metric space $T$ onto $X$ such that $$d(f^{-1}(p),f^{-1}(q))>0,$$ for distinct points $p, q \in X.$

The author difines $T$ as follows:

$T=\lbrace (\alpha_1,\alpha_2,...)\in N(A): \bigcap \lbrace U_{\alpha_n}^n: n\in N\rbrace\not=\emptyset \rbrace$, where $\lbrace \mathcal U_n=\lbrace U_{\alpha}^n: \alpha \in A, n \in N\rbrace$ is a sequence of open covering of $X$ satisfying the condition in Lemma 1. (it can be seen in the paper.)

The author difines $f: T \rightarrow X$ as follows:

$f(\alpha)=\bigcap \lbrace U_{\alpha_n}^n: n\in N \rbrace$ for $\alpha \in T$

My question is this:

1) What is the topology the author used which make $T$ is metrizable?

2) Is $f$ continuous?

Thanks for your help.

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  • $\begingroup$ As for the first question, isn't it answered in the paper? Otherwise you rather ask the author. $\endgroup$ – Fernando Muro Apr 26 '13 at 7:10
  • $\begingroup$ The author said unclearly in the paper. $\endgroup$ – Paul Apr 26 '13 at 7:16
  • $\begingroup$ Did you write her/him? I'd say it's be best solution. Even from my point of view it should be a step previous to publicly post a question like this. $\endgroup$ – Fernando Muro Apr 26 '13 at 7:34
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    $\begingroup$ The author mentions in the paper that he uses Baire’s zero-dimensional metric space N(A). Baire space ( en.wikipedia.org/wiki/Baire_space_%28set_theory%29 ) is well-known in the descriptive set theory, it has a basis consisting of sets of all sequences with prescribed first $n$ elements and we can get a metric by putting $d(x,y)=1/\min\{n; x_n\ne y_n\}$. I have seen it so far only for sequences of integers, but using other set $A$ instead of integers probably does not make much difference. $\endgroup$ – Martin Sleziak Apr 26 '13 at 8:05
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    $\begingroup$ I don’t think $f$ needs to be continuous in general. Fix $X$ and $\mathcal U_n$, and construct $T$ and $f$ as in the proof. Then if we take $Y$ to be a finer topology than $X$ on the same underlying set, the same $\mathcal U_n$ can be used for $Y$, yielding the same $T$ and $f$. In particular, $Y$ can be taken discrete. Thus, if the construction as given always yielded a continuous $f$, the point preimages $f^{-1}(p)$ would need to be open (even clopen), and this does not appear to be the case. $\endgroup$ – Emil Jeřábek Apr 26 '13 at 13:31

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