nLab
G-delta subspace

Definition

Definition

A G-delta, $G_{\delta }$, subset of a topological space is a set that can be written as the intersection of a countable family of open sets.

Results

One place where $G_{\delta }$-subsets occur is when looking at continuous maps from an arbitrary topological space to a metric space (or, more generally, a first countable space). In particular, when considering continuous real-valued functions. Thus we have the following connections to the separation axioms.

Theorem

A normal space in which every closed set is a $G_{\delta }$-set is perfectly normal?.

Theorem

In a completely regular space, every singleton set that is a $G_{\delta }$-set is the unique global maximum of a continuous real-valued function.

Proof

One direction is obvious. For the other, let $v$ be a point in a completely regular space $X$ such that $\{v\}$ is a $G_{\delta }$-set. Let $\{V_n\}$ be a sequence of open sets such that $\bigcap V_n=\{v\}$. We now define a sequence of functions $(f_n)$ recursively with the properties:

1. $f_n:X\to [0,1]$ is a continuous function,
2. $f_n(v)=1$,
3. $f_n^{-1}(1)$ is a neighbourhood of $v$,
4. for $n>1$, $f_n$ has support in $V_n\cap f_{n-1}^{-1}(1)$ whilst $f_1$ has support in $V_1$,

Having defined $f_1,\dots ,f_{n-1}$, we define $f_n$ as follows. Since $V_n\cap f_{n-1}^{-1}(1)$ is a neighbourhood of $v$ and $X$ is completely regular, there is a continuous function ${\tilde{f}}_n:X\to [0,1]$ with support in this neighbourhood and such that ${\tilde{f}}_n(v)=1$. We then compose with a continuous, increasing surjection $[0,1]\to [0,1]$ which maps $[\frac{12}{,}1]$ to $1$. The resulting function is the required $f_n$.

We then define a function $f:X\to [0,1]$ by

$$f(x)≔\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}{f}_n(x).$$f(x) \coloneqq \sum_{n=1}^\infty \frac1{2^n} f_n(x).

By construction, $f^{-1}(1)=\{v\}$.

We need to prove that this is continuous. First, note that if $f_n(x)\ne 0$ then $f_k(x)=1$ for $k<n$ and if $f_n(x)\ne 1$ then $f_k(x)=0$ for $k>n$. Hence the preimage under $f$ of $(\frac{2^k-1}{2^k},\frac{2^{k+1}-1}{2^{k+1}})$ is $f_n^{-1}(0,1)$ and $f$ restricted to this preimage is a scaled translate of $f_n$. From this, we deduce that the preimage of any open set not containing $1$ is open. Thus $f$ is continuous everywhere except possibly at $v$. Continuity at $v$ is similarly simple: given a set of the form $(1-ϵ,1]$ then there is some $n$ such that $2^{-n}<ϵ$, whence $f^{-1}(1-ϵ,1]$ contains all points such that $f_k(x)=1$ for $k\le n$, which by construction is a neighbourhood of $v$. Hence $f$ is continuous and has a single global maximum at $v$.