Open Sets

Definition: Open Subset

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

A subset of $X$ is open iff it is an element of $\tau$.

Openness Criteria

THEOREM

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

A subset $U\subseteq X$ is open if and only if each $x\in U$ has a neighbourhood $N(x)$ such that $N(x)\subseteq U$.

PROOF

If $U$ is open, then $U$ is, by definition, a neighbourhood of every $x\in U$.

If there exists a neighbourhood $N(x)$ of each $x$ such that $N(x)\subseteq U$, then inside each neighbourhood $N(x)$ there exists, by definition, an open set $O(x)\subseteq N(x)$ which contains $x$. Since $N(x)\subseteq U$, we have $O(x)\subseteq N(x)\subseteq U$. This means that we can construct $U$ as the union of these open sets $O(x)$.

$$U=\bigcup _{x\in U}O(x)$$

Since this is a union of open sets, i.e. it is a union of a subset of the topology $\tau$, we have that $U\in \tau$.

Theorem

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

A subset $U\subseteq X$ is open if and only if for each $x\in U$ there exists an open set $V$ such that $V\subseteq U$ and $x\in V$.

PROOF

TODO

Theorem

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

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

$$S=\mathrm{int}S$$

PROOF

We need to prove two things:

• (I) If $S$ is open, then $S=\mathrm{int}S$.
• (II) If $S=\mathrm{int}S$, then $S$ is open.

Proof of (I):

Suppose $S$ is open. Recall the definition of the interior $\mathrm{int}S$:

$$\mathrm{int}S\stackrel{\text{def}}{=}\bigcup \{O\subseteq S\mid O\text{ is open}\}$$

Since $S\subseteq S$ and $S$ is open, we know that $S\in \{O\subseteq S\mid O\text{ is open}\}$ and thus $S\subseteq \mathrm{int}S$. However, the interior is a subset of $S$. Since $S\subseteq \mathrm{int}S$ and $\mathrm{int}S\subseteq S$, we know deduce that $S=\mathrm{int}S$.

Proof of (II):

Suppose that $S=\mathrm{int}S$. Since the interior $\mathrm{int}S$ is a union of open sets, it is itself open. Therefore, $S$ is open.

Properties

Theorem: Union of Open Sets

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

The union of any collection of open subsets is also open.

PROOF

This follows directly from the definition of a topology.

Theorem: Intersection of Open Sets

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

The intersection of any finite collection of open sets is also open.

PROOF

We consider $n\ge 2$ arbitrary open subsets $U_1,\cdots ,U_n$.

For $n=2$, the definition of the topology $\tau$ tells us that $U_1\cap U_2\in \tau$.

Now suppose $U^{\prime }=U_1\cap \cdots \cap U_{n-1}\in \tau$. We have

$$\begin{array}{rl}{U}_1\cap \cdots \cap U_n& =(U_1\cap \cdots \cap U_{n-1})\cap U_n\\ U_1\cap \cdots \cap U_n& =U^{\prime }\cap U_n\end{array}$$

Since $U_1\cap \cdots \cap U_n$ is the intersection of two elements of $\tau$, it must itself be an element of $\tau$, Q.E.D.