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 $$x \in S'$$. $$\qquad \square$$

Next maths post Previous maths post

All maths posts