The Interior of Open Sets in a Topological Space

# The Interior of Open Sets in a Topological Space

Recall from The Interior Points of Sets in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $a \in A$ is said to be an interior point of $A$ if there exists an open neighbourhood $U$ of $a$ contained in $A$, that is, there exists a $U \in \tau$ with $a \in U$ such that:

(1)
\begin{align} \quad a \in U \subseteq A \end{align}

We called the set of all interior points of $A$ the interior of $A$ and denoted it by $\mathrm{int} (A)$. Equivalently, we saw that $\mathrm{int} (A)$ is the largest open set contained in $A$.

We will now look at a very useful theorem which will tell us that a set is open if and only if $A = \mathrm{int} (A)$.

 Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is open if and only if every $a \in A$ is an interior point of $A$, i.e., $A = \mathrm{int} (A)$.
• Proof: $\Rightarrow$ Suppose that $A$ is an open set. Then $A \in \tau$ and so for every element $a \in A$ let $U = A \in \tau$ so that:
(2)
\begin{align} \quad a \in U = A \subseteq A \end{align}
• Hence every point $a \in A$ is an interior point of $A$.
• $\Leftarrow$ Now suppose that every $a \in A$ is an interior point of $A$. Then for every $a \in A$ there exists a an open subset $U_a \in \tau$ such that:
(3)
\begin{align} \quad a \in U_a \subseteq A \end{align}
• So $\displaystyle{A = \bigcup_{a \in A} U_a}$ with is a union of arbitrary open sets which is open. Therefore $A$ is open. $\blacksquare$