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

*

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


*Is $f$ continuous?
Thanks for your help.
 A: 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.
