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 \(X \backslash S\).


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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