G-delta subspace

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


One place where G δG_\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.


A normal space in which every closed set is a G δG_\delta-set is perfectly normal.


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


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

  1. f n:X[0,1]f_n \colon X \to [0,1] is a continuous function,
  2. f n(v)=1f_n(v) = 1,
  3. f n 1(1)f_n^{-1}(1) is a neighbourhood of vv,
  4. for n>1n \gt 1, f nf_n has support in V nf n1 1(1)V_n \cap f_{n-1}^{-1}(1) whilst f 1f_1 has support in V 1V_1,

Having defined f 1,,f n1f_1, \dots, f_{n-1}, we define f nf_n as follows. Since V nf n1 1(1)V_n \cap f_{n-1}^{-1}(1) is a neighbourhood of vv and XX is completely regular, there is a continuous function f˜ n:X[0,1]\tilde{f}_n \colon X \to [0,1] with support in this neighbourhood and such that f˜ n(v)=1\tilde{f}_n(v) = 1. We then compose with a continuous, increasing surjection [0,1][0,1][0,1] \to [0,1] which maps [12,1][\frac12,1] to 11. The resulting function is the required f nf_n.

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

f(x) n=1 12 nf n(x). f(x) \coloneqq \sum_{n=1}^\infty \frac1{2^n} f_n(x).

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

We need to prove that this is continuous. First, note that if f n(x)0f_n(x) \ne 0 then f k(x)=1f_k(x) = 1 for k<nk \lt n and if f n(x)1f_n(x) \ne 1 then f k(x)=0f_k(x) = 0 for k>nk \gt n. Hence the preimage under ff of (2 k12 k,2 k+112 k+1)(\frac{2^k-1}{2^k}, \frac{2^{k+1}-1}{2^{k+1}}) is f n 1(0,1)f_n^{-1}(0,1) and ff restricted to this preimage is a scaled translate of f nf_n. From this, we deduce that the preimage of any open set not containing 11 is open. Thus ff is continuous everywhere except possibly at vv. Continuity at vv is similarly simple: given a set of the form (1ϵ,1](1 -\epsilon,1] then there is some nn such that 2 n<ϵ2^{-n} \lt \epsilon, whence f 1(1ϵ,1]f^{-1}(1-\epsilon,1] contains all points such that f k(x)=1f_k(x) = 1 for knk \le n, which by construction is a neighbourhood of vv. Hence ff is continuous and has a single global maximum at vv.

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