# nLab G-delta subspace

topology

algebraic topology

# Contents

## 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 $\left\{v\right\}$ is a ${G}_{\delta }$-set. Let $\left\{{V}_{n}\right\}$ be a sequence of open sets such that $\bigcap {V}_{n}=\left\{v\right\}$. We now define a sequence of functions $\left({f}_{n}\right)$ recursively with the properties:

1. ${f}_{n}:X\to \left[0,1\right]$ is a continuous function,
2. ${f}_{n}\left(v\right)=1$,
3. ${f}_{n}^{-1}\left(1\right)$ is a neighbourhood of $v$,
4. for $n>1$, ${f}_{n}$ has support in ${V}_{n}\cap {f}_{n-1}^{-1}\left(1\right)$ whilst ${f}_{1}$ has support in ${V}_{1}$,

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

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

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

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

We need to prove that this is continuous. First, note that if ${f}_{n}\left(x\right)\ne 0$ then ${f}_{k}\left(x\right)=1$ for $k and if ${f}_{n}\left(x\right)\ne 1$ then ${f}_{k}\left(x\right)=0$ for $k>n$. Hence the preimage under $f$ of $\left(\frac{{2}^{k}-1}{{2}^{k}},\frac{{2}^{k+1}-1}{{2}^{k+1}}\right)$ is ${f}_{n}^{-1}\left(0,1\right)$ 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 $\left(1-ϵ,1\right]$ then there is some $n$ such that ${2}^{-n}<ϵ$, whence ${f}^{-1}\left(1-ϵ,1\right]$ contains all points such that ${f}_{k}\left(x\right)=1$ for $k\le n$, which by construction is a neighbourhood of $v$. Hence $f$ is continuous and has a single global maximum at $v$.

Revised on June 22, 2010 23:27:04 by Toby Bartels (75.88.99.206)