The characteristic function and the Sierpinski-space

9 Juin 20235 Juin 2023 / Christian Nguembou Tagne

A topological space, whose underlying set has two elements, is called the Sierpinski-space if its topology is non-discrete and non-indiscrete. In this publication, we give a necessary and sufficient condition for the continuity of a characteristic function from a topological space into the Sierpinski-space.

In fact, this is our solution to an exercise outlined in a previous entry of this blog. The statement of the exercise in question is repeated here for the comfort of reading.

Exercise.

We set $2=\bigl\{0,1\bigr\}$ , and recall that the characteristic function of a subset $A$ of a set $X$ is the function denoted by $\chi_{A}$ , from $A$ into $2$ , defined by

$\chi_{A}(x)=\left\{\begin{array}{l} 1\,\text{ if }\, x\in A, \\[6pt] 0\,\text{ if }\, x\in X\setminus A.\end{array}\right.$

Verify that the function $\gamma:\mathcal{P}(X)\rightarrow 2^{X}$ , from the power set of $X$ into the set of all functions of $X$ into $2$ , given by $\gamma(A)=\chi_{A}$ , is a bijection.
Given a set $X$ and a subset $A$ of $X$ , show that the set $\mathfrak{T}=\bigl\{\emptyset, A,X\bigr\}$ is a topology on $X$ . In particular, the set $\mathfrak{A}=\bigl\{\emptyset,\{1\},2\bigr\}$ is a topology on $2$ . The topological space $(2,\mathfrak{A})$ is called the Sierpinski-space.
Let $(X,\mathfrak{O})$ be a topological space. Prove that the characteristic function $\chi_{A}:(X,\mathfrak{O})\rightarrow(2,\mathfrak{A})$ of a subset $A$ of $X$ is continuous if, and only if, $A\in\mathfrak{O}$ .