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Hello, there is a statement as following:

If every point of X is a G_delta and X is T_1, then take Y = set of X, plus the topology generated by all open sets needed to prove G_delta-ness of every singleton, plus the cofinite topology, then Y is a condensation of X (using identity) and is first countable by construction.

So a T_1 space X has a first countable T_1 condensation iff every point is a G-delta

I have some questions about this statement which need to be solved.

1, why the space Y is first countable?

2, if we delete the cofinite topology, the space Y is still the first countable T_1 condensation?

3, how to make the topology generated on Y be first countable and hausdorff, if X is hausdorff and

every point of X is a G_delta.

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Cofinite is not needed, T_1 follows regardless – Henno Brandsma Oct 12 '11 at 8:57

2 Answers 2

up vote 2 down vote accepted

So we start with $(X, \mathcal{T})$, a $T_1$ space in which every point is a $G_{\delta}$, as witnessed by open sets $U_n(x)$, $n \in \mathbb{N}$, $x \in X$. W.l.o.g. we can take these sets to be decreasing.

The topology generated by all these sets we call $\mathcal{T}'$, say, and it is $T_1$, because for every $ x \neq y $, there is some $U_n(x)$ that does not contain $y$ (or else $y$ would be in their intersection, for all $n$, and this intersection is precisely $\{x\}$), and this witnesses the $T_1$ property ($\{y\}$ is closed, by this argument).

A base for this topology is given by all finite intersections of the sets $U_n(x)$. It's not clear how this topology would be first countable; the obvious candidate for a local base at $x$ is all sets $U_n(x)$, but if $z$ is in $U_n(x) \cap U_m(y)$, I don't see how we automatically get some $U_m(z)$ inside it. So in general, this might not work as advertised.

This might be remedied by using elementary submodels (see e.g. I think. This would also then do the Hausdorff case, which this construction (even if it could be made to work) does not always give, e.g. if we take a countable nowhere first countable space and use the cofinite topology on it. This is a $T_1$ condensation obtained by the described construction, which is not Hausdorff.

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There is a Hausdorff space with all points $G_\delta$ but without a condensation to a first countable Hausdorff space.

Let $X$ be the space of ultrafilters on $\mathbb{N}$ with the topology generated by sets of the form $\{\mathcal{p}\}\cup A$ where $A\in p\in X$ and I have committed the common abuse of notation that identifies $\mathbb{N}$ with the principal ultrafilters on $\\mathbb{N}$. This makes $X$ Hausdorff (but not regular). Moreover, every point is $G_\delta$ because $X$ is $T_1$ and every point has a countable neighborhood. Crucially, $X$ has cardinality $2^{2^{\aleph_0}}$.

Now suppose $Y$ is the set $X$ with a coarser first countable Hausdorff topology. For each $p\in X$, let $(U_n(p):n\in\mathbb{N})$ be a base of $Y$-neighborhoods of $p$; let $V_n(p)=U_n(p)\cap\mathbb{N}$. Because $Y$ is Hausdorff and $\mathbb{N}$ is $X$-dense, we have $\vec{V}(p)\not=\vec{V}(q)$ for all distinct $p,q$. Therefore, $X$ injects into $\mathcal{P}(\mathbb{N})^{\mathbb{N}}$. But $X$ is too big, so actually there is no such $Y$.

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