Closed Sets

Definition: Closed Set

Let $(S,\tau )$ be a topological space.

A subset $U$ of $S$ is closed if its complement $S\setminus U$ is an open set.

THEOREM

Let $(S,\tau )$ be a topological space.

A subset $C\subseteq S$ is closed if and only if for each $x$ in its complement $S\setminus C$ there exists a neighbourhood $N$ of $x$ such that $N\subseteq S\setminus C$.

PROOF

TODO

Closedness Criteria

THEOREM

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is closed if and only if it is equal to its own closure.

PROOF

We have to prove two things:

• (I) If $S$ is closed, then $S=\overline{S}$.
• (II) If $S=\overline{S}$, then $S$ is closed.

Proof of (I):

TODO

Proof of (II):

TODO

Theorem: Limit Points of Closed Subsets

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is closed if and only if it contains all of its limit points.

PROOF

We need to prove two things separately:

• (I) If $S$ is closed, then every limit point of $S$ lies in $S$.
• (II) If $S$ contains all of its limit points, then it is closed.

Proof of (I):

We prove this by contradiction. Suppose that $S$ is closed and there exists a limit point $p$ of $S$ which lies outside of $S$, i.e. $p\in X\setminus S$. We know that $X\setminus S$ is open because it is the complement of a closed set. However, since $p$ is a limit point of $S$, every open set which contains $p$ must also contain an element of $S$. This implies that $X\setminus S$ contains an element of $S$ which is a contradiction.

Proof of (II):

Suppose that $S$ contains all of its limit points. This means that there are no points $p\in X\setminus S$ such that every open set $U$ which contains $p$ also contains another element of $S$. Alternatively, this means that each $p\in X\setminus S$ is contained in some open set $U_p$ such that $U_p\cap S=\varnothing$, i.e. $U_p\subseteq X\setminus S$. Therefore, the union ${\bigcup }_{p\in X\setminus S}{U}_p$ is a subset of $X\setminus S$ and, since it contains every $p\in X\setminus S$, it means that ${\bigcup }_{p\in X\setminus S}{U}_p=X\setminus$. Since $X\setminus S$ is a union of open sets, it is itself open and so $S$ is closed.

Properties

Theorem: Intersection of Closed Sets

Let $(S,\tau )$ be a topological space.

The intersection of any collection of closed subsets is also closed.

PROOF

TODO

Theorem: Union of Closed Sets

Let $(S,\tau )$ be a topological space.

The union of any finite collection of closed subsets is also closed.

PROOF

Let $S_1,\cdots ,S_n$ be closed sets. We need to show that $\bigcup \{S_1,\cdots ,S_n\}=S_1\cup \cdots \cup S_n$ is closed. According to the definition of a closed set, this means we must show that $S\setminus (S_1\cup \cdots \cup S_n)$ is open.

By the distributive law

$$S\setminus (S_1\cup \cdots \cup S_n)=(S\setminus S_1)\cap \cdots \cap (S\setminus S_n)$$

The sets $S_1,\cdots ,S_n$ are closed and by definition their complements $S\setminus S_1,\cdots ,S\setminus S_n$ are open. The right-hand side is thus an intersection of open sets and is, therefore, open itself. This means that the complement $S\setminus S_1\cup \cdots \cup S_n$ is open and so $S_1\cup \cdots \cup S_n$ is closed.