A characterization of the existence of an interior point

We recall that, in a topological space, a point is said to be an interior point of a subset if the latter is a neighborhood of the point in question. The set of interior points of a subset is called the interior of this subset. It is well-known that a subset is open if and only if it coincides which its interior.

In this note, we prove that the interior of a subset of a topological space contains at least one point if and only if the subset in question meets each dense subset of the topological space. In fact, this text is a solution to an exercise stated in a previous contribution.

Exercise.

Prove that a subset $A$ of a topological space $X$ meets each dense subset of $X$ if and only if the interior of $A$ is not empty.

Solution of the Exercise

First, let $A$ meet each dense subset of $X$ . We assume that the interior $i(A)$ of $A$ is empty. Then $\overline{X\setminus A}=X\setminus i(A)=X\setminus\emptyset=X$ , that is, $X\setminus A$ is a dense subset of $X$ . By the hypothesis, this yields $A\cap(X\setminus A)\neq\emptyset$ : a contradiction. The assumption $i(A)=\emptyset$ is thus false. Hence $i(A)\neq\emptyset$ .

Conversely, let $i(A)$ be non-empty. Moreover, let $D$ be a dense subset of $X$ . Then $i(A)\cap D\neq\emptyset$ , since $i(A)$ is an open set. (Recall that a subset $D$ of a topological space $X$ is said to be dense if $\overline{D}=X$ , that is, if every non-empty open set meets $D$ .) Since $i(A)\cap D\subset A\cap D$ , it follows that $A\cap D\neq\emptyset$ . Therefore $A$ meets each dense subset of $X$ .

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Formalis Mathematica

Rigueur et formalisme au cœur de l’apprentissage et de l’enseignement des mathématiques

A characterization of the existence of an interior point

Exercise.

Solution of the Exercise

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Exercise.

Solution of the Exercise

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