# Interiors, closures and boundaries of topological spaces

I’d show alternate definitions of interiors, closures and boundaries for topological spaces which I’ve found immensely helpful.

Let’s go through the usual definitions first. Given a topological space \(X\) with a subset \(S \subset X\), the interior \(\text{int}(S)\) of \(S\) is defined to be the union of all open subsets of \(X\) that is contained in \(S\). The closure \(\overline{S}\) is defined to be the intersection of all closed subsets of \(X\) containing \(S\). \(\overline{S} \backslash \text{int}(S)\) is known as the boundary of \(S\) and denoted \(\delta S\).

The equivalent definitions I like to use are

\[\begin{align*} a \in \text{Clo}(S) \iff& \nexists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset X \backslash S\\ a \in \text{Int}(S) \iff& \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset S\\ a \in \delta S \iff&\nexists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset S \\ &\text{ and } \nexists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset X \backslash S \end{align*}\]The similarities in the definitions prompts us to define the following. Given some \(a \in X\) if we let \(P\) and \(Q\) be the statements

\[\begin{align*} P &: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset S\\ Q &: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset X \backslash S \end{align*}\]It’s impossible for \(P\) and \(Q\) to both be true so we can have the truth table

This means \(\text{Int}(S)\), \(\delta S\) and \(\text{Int}(X \backslash S)\) partition \(X\)

## Accumulation and isolated points

Let \(S'\) denote the set of accumulation points of \(S\) and \(\text{Iso}(S)\) denote the set of isolated points of \(S\), we have the equivalent definitions

\[\begin{align*} a \in S' &\iff \nexists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \backslash \{a\} \subset X \backslash S \\ a \in \text{Iso}(S) &\iff \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \backslash \{a\} \subset X \backslash S\text{ and } a \in S \\ \end{align*}\]This shows we’re concerned with \(U \backslash \{a\}\). Inspired from how we defined \(P\) and \(Q\), it prompts us to define the following statements \(V\), \(W\) and \(K\) for some \(a \in X\) as follows.

\[\begin{align*} P &: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset S\\ Q &: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \subset X \backslash S \\ V&: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \backslash \{a\} \subset S \\ W&: \exists \text{ open } U \subset X \text{ s.t. } a \in U \text{ and } U \backslash \{a\} \subset X \backslash S \\ K&: a \in S \end{align*}\]Using this language, we can see the following equivalences.

\[\begin{align*} a \in S' &\iff \neg W \\ a \in \text{Iso}(S) &\iff W \land K \end{align*}\]Also note that

\[\begin{align*} P &\iff V \land K \\ Q &\iff W \land \neg K \end{align*}\]As such \(\neg Q \iff \neg W \lor K \iff (\neg W) \text{ XOR } (W \land K)\), so accumulation points and isolated points form a partition of the closure.

We can further split into cases. Considering the interior of \(S\) first, notice that \(P \land \neg Q \iff V \land K\). So we have the following cases

Considering the boundary \(\delta S\), we also have \(\neg P \land \neg Q \iff (K \land \neg V) \lor (\neg K \land \neg W)\). So we have

We can conduct a similar analysis for the interior of \(X \backslash S\).

## Examples

Consider the real line with the topology induced from the Euclidean metric. Let \(S=((0,2] \cup \{3\})\backslash\{1\}\). Then we have

If we consider the discrete topological space \(X\), then every point of a subset \(S\) is an interior and isolated point of \(S\).

## T1 Spaces

A topological space \(X\) is \(T_1\) if for every pair of distinct points, each has a neighbourhood not containing the other point. This is equivalent to the statement that for all elements \(x\) of \(X\), \(\{ x \}\) is closed in \(X\).

As such if \(X\) is \(T_1\) then for any open \(U\) in \(X\) and \(x\) in \(X\), \(U \cap (X \backslash \{x\})\) is open. I.e. we can take away a finite number of points from an open set and it would still be open.

**Claim:** If \(X\) is \(T_1\), then for any element \(x \in S'\), any open \(U \text{ s.t. } x \in U\) contains infinitely many points of \(S\).

**Proof:** Suppose for contradiction that there exists some neighbourhood of \(x\) that has a finite number of intersections with \(S\). Let those points be \(x_1, x_2, \dots x_n\). Then consider \(U \backslash \{x_1, \dots x_n\}\), the set is still open by \(X\) hausdorff and \(U \backslash \{x_1, \dots x_n\} \backslash \{x\} \subset X \backslash S\). This contradicts \(x \in S'\). \(\qquad \square\)

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