# 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 $\{v\}$ 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. $f_n \colon X \to [0,1]$ is a continuous function,
2. $f_n(v) = 1$,
3. $f_n^{-1}(1)$ is a neighbourhood of $v$,
4. for $n \gt 1$, $f_n$ 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 $\tilde{f}_n \colon X \to [0,1]$ 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 $[\frac12,1]$ to $1$. The resulting function is the required $f_n$.

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

$f(x) \coloneqq \sum_{n=1}^\infty \frac1{2^n} f_n(x).$

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 $k \lt n$ and if $f_n(x) \ne 1$ then $f_k(x) = 0$ for $k \gt n$. 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 $(1 -\epsilon,1]$ then there is some $n$ such that $2^{-n} \lt \epsilon$, whence $f^{-1}(1-\epsilon,1]$ 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 $v$.

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