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 defines $T$ as follows:

$T=\{(\alpha_1,\alpha_2,...)\in N(A): \bigcap \{U_{\alpha_n}^n: n\in N\}\not=\emptyset\}$, where $\{\mathcal U_n=\{U_{\alpha}^n: \alpha \in A, n \in N\}$ 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:

$f(\alpha)=\bigcap \{U_{\alpha_n}^n: n\in N\}$ 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.

  • $\begingroup$ I think you should have mentioned that you have posted the same question at mathoverflow, too. $\endgroup$ – Martin Sleziak Apr 26 '13 at 8:10
  • 1
    $\begingroup$ The author mentions in the paper that he uses Baire’s zero-dimensional metric space N(A). The Baire space 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. This is precisely the product topology on $A^\omega$, where $A$ has discrete topology. $\endgroup$ – Martin Sleziak Apr 26 '13 at 8:14
  • $\begingroup$ I am not sure what is the usual protocol for questions posted to both MSE and MO. But I believe that if you will find my answer (or any other answer, if more answers will be given here) satisfying; you should probably at least mention at MO that you have already received some answer here. (Of course, it is still possible that you will get some more interesting answer there.) $\endgroup$ – Martin Sleziak Apr 26 '13 at 21:17

The author mentions in the paper that he uses Baire’s zero-dimensional metric space N(A).

The Baire space is well-known in the descriptive set theory, it is usually defined on the set $\omega^\omega$ of all integer sequences.

In the case of this paper the author obviously works with (a subspace of) the space $A^\omega$ of all sequences of elements from $A$.

The following possibilities give the same topology on $A^\omega$:

  • The product topology, where $A$ has the discrete topology.
  • The metric given by $d(x,y)=1/\min\{n; x_n\ne y_n\}$.
  • The topology generated by the base consisting of all sets $N(\alpha_1\dots\alpha_n)=\{x\in A^\omega; x_1=\alpha_1,\dots,\alpha_n\}$; i.e., the basic sets consists of all sequences with prescribed first $n$ elements.

If you check the paper then you see that the author uses precisely the basic sets of this form when he verifies whether the map $f$ is open.

This is a slight digression from the question, but it might be interesting for you. (But you can ignore this part if you prefer - it is not necessary for you when reading this paper; I am only mentioning this to stress the similarity with another commonly used construction.)

All this can be nicely visualized using trees. A good reading on this topic is the chapter on trees in Kechris' Classical Descriptive Set Theory.

You can read there about Lusin scheme, which is a construction very similar to the construction of the function $f$ in this paper.

In particular, the construction of a map associated to a Lusin scheme (in Proposition 7.6) is very similar to the definition of the function in your paper. The difference is that the target space is a metric space and that we assume that the sets have decreasing diameters (instead of the assumption that they come from some covers; as in your paper). In Proposition 7.6 it is shown that this map is continuous.

Back to your original question.

Now if you want to show that the map $f$ is not necessarily continuous, take any space $X$ which is submetrizable but not metrizable. Let $d$ be a metric on $X$, which yields the coarser metrizable topology.

Take the covers $\mathcal U_n=\{B(x,1/n); x\in X\}$, where $B(x,r)=\{y\in X; d(x,y)<r\}$ are the balls w.r.t. the metric $r$. These covers fulfill the assumptions from the paper.

Now since your topology is strictly finer than the topology given by the metric $d$, there exists a point $x$ and an open set $U\ni x$ such that no $d$-ball around $x$ lies entirely in the set $U$.

The constant sequence $\overline x=(x,x,x,\dots)$ is mapped to $x$ by the map $f$ defined in the paper.

Basic neighborhoods of the point $\overline x$ are of the form $N(\underset{\text{$n$-times}}{\underbrace{x,x,\dots,x})}$; this set contains all sequences where the first $n$ coordinates are equal to $x$.

Now if you take a point $x_n\in B(x,1/n)\setminus U$ and make the sequence $\overline x_n =(x,x,\dots,x,x_n,x_n,\dots)$; then this sequence belongs to same basic neighborhood and it is mapped to the point $x_n\notin U$. The assumption $d(x,x_n)$ implies that the intersection of the sets $U^k_\alpha$ contains the point $x_n$ and thus it is non-empty (this is the condition which defines the subspace $T$ in the paper; here $U^k_\alpha=B(x,1/k)$ for $k\le n$ and $U^k_\alpha=B(x_n,1/k)$ for $k>n$.)

So it is not possible to find a basic neighborhood $N$ such that $f[N]\subseteq U$. This contradicts the continuity of $f$.

EDIT: In your comment below you are asking what the set $A$ is.

Again I quote from the paper:

where $\{\mathcal U_n=\{U^n_\alpha; \alpha\in A\}; n\in N\}$ is a sequence of open coverings of $X$ satisfying the condition in Lemma 1.

Lemma 1 looks like this:

Lemma 1. A space $X$ has a $G_\delta$-diagonal iff (=if and only if) there is a sequence $\{\mathcal U_n; n\in N\}$ of open coverings of $X$ such that for each point $p$ in $X$ $$p=\bigcap \{S(p,\mathcal U_n; n\in N)\}.$$

From the notation in the proof it seems that the author additionaly assumes that the open covers $\mathcal U_n$ are indexed by the same set $A$. This can be obtained by taking union of their index sets and adding arbitrary open sets on the new indices (for example empty sets).

Brian M. Scott gave you a similar suggestion in his comment here.

  • $\begingroup$ Thanks for you answer. Sorry I still don't know what is $A$ when the author defines $T$. $\endgroup$ – Paul Apr 27 '13 at 0:39
  • $\begingroup$ @Paul I've added to my answer how I understand the set $A$. (I hope that if I'm mistaken, someone will correct me; but this seems to be similar to Brian M. Scott's comment from the other thread; which increases probability that it is correct.) $\endgroup$ – Martin Sleziak Apr 27 '13 at 5:16

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