Basic Thms. Regarding the Interior Points of Sets in a Top. Space

# Basic Theorems Regarding the Interior Points of Sets in a Topological Space

Recall from the 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 called an interior point of $A$ if there exists an open neighbourhood $U \in \tau$ such that $a \in U \subseteq A$. The set of all interior points of $A$ is denoted as $\mathrm{int} (A)$.

We have already proved some results regarding interior points of a set $A \subseteq X$ for a topological space $(X, \tau)$ including:

• $A$ is open if and only if $A = \mathrm{int} (A)$.
• If $U$ is an open set such that $U \subseteq A$ then $U \subseteq \mathrm{int} (A)$.
• $\mathrm{int} (A)$ is the largest open subset of $A$.

We will now look at some more basic theorems regarding the interior points of sets in a topological space.

 Theorem 1: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. If $A \subseteq B$ then $\mathrm{int} (A) \subseteq \mathrm{int} (B)$.
• Proof: Suppose that $A \subseteq B$. Consider the set $\mathrm{int} (A)$. Then $\mathrm{int} (A)$ is the largest open subset of $A$ so $\mathrm{int} (A) \subseteq A$. But then $\mathrm{int} (A)$ is an open subset of $B$, which must be contained in the largest open subset of $B$ which is $\mathrm{int} (B)$. Hence:
(1)
 Theorem 2: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\mathrm{int} (A) \cup \mathrm{int} (B) \subseteq \mathrm{int} (A \cup B)$.
• Proof: Let $x \in \mathrm{int} (A) \cup \mathrm{int} (B)$. Then $x \in \mathrm{int} (A)$ or $x \in \mathrm{int} (B)$.
• If $x \in \mathrm{int} (A)$ then there exists a $U \in \tau$ such that $x \in U \subseteq A$. So $x \in U \subseteq A \cup B$, so $x \in \mathrm{int} (A \cup B)$. Similarly, if $x \in \mathrm{int} (B)$ then there exists a $U \in \tau$ such that $x \in U \subseteq B$. So $x \in U \subseteq A \cup B$, so $x \in \mathrm{int} (A \cup B)$. Hence:
(2)
\begin{align} \quad \mathrm{int} (A) \cup \mathrm{int} (B) \subseteq \mathrm{int} (A \cup B) \quad \blacksquare \end{align}

Important Note!

It is important to note that in general $\mathrm{int} (A) \cup \mathrm{int} (B) \not \supseteq \mathrm{int} (A \cup B)$. For example, consider the sets $A = (0, 1]$ and $B = (1, 2]$. Then $\mathrm{int} (A) = (0, 1)$ and $\mathrm{int} (B) = (1, 2)$ so $\mathrm{int} (A) \cup \mathrm{int} (B) = (0, 1) \cup (1, 2) = (0, 2) \setminus \{ 1 \}$.

Furthermore, $A \cup B = (0, 2]$ and $\mathrm{int} (A \cup B) = (0, 2)$.

We clearly see that $\mathrm{int} (A) \cup \mathrm{int} (B) \not \supseteq \mathrm{int} (A \cup B)$.

For another example, consider the set $X = \{ a, b, c, d \}$ and the topology $\tau = \{ \emptyset, \{ b \} \{ a, b \}, \{b, c \}, \{a, b, c \}, X \}$.

Now consider the set $A = \{ a \}$ and $B = \{ b, c \}$. Then $\mathrm{int} (A) = \emptyset$ and $\mathrm{int} (B) = B$. So $\mathrm{int} (A) \cup \mathrm{int} (B) = \emptyset \cup B = B$. However $A \cup B = \{ a, b, c \}$ and $\mathrm{int} (A \cup B) = A \cup B$. Clearly $\mathrm{int} (A) \cup \mathrm{int} (B) = B \subseteq A \cup B = \mathrm{int} (A \cup B)$, however $\mathrm{int} (A) \cup \mathrm{int} (B) \not \supseteq \mathrm{int} (A \cup B)$.

 Theorem 3: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\mathrm{int} (A) \cap \mathrm{int} (B) = \mathrm{int} (A \cap B)$.
• Proof: Let $x \in \mathrm{int} (A) \cap \mathrm{int} (B)$. Then $x \in \mathrm{int} (A)$ and $x \in \mathrm{int} (B)$. Since $x \in \mathrm{int} (A)$ there exists a $U_A \in \tau$ such that $x \in U_A \subseteq A$. Similarly, since $x \in \mathrm{int} (B)$ there exists a $U_B \in \tau$ such that $x \in U_B \subseteq B$. Let $U = U_A \cap U_B$. Since $U_A \in \tau$ and $U_B \in \tau$ we have that $U = U_A \cap U_B \in \tau$.
• So $x \in U_A \cap U_B \subseteq A \cap B$, so $x \in \mathrm{int} (A \cap B)$ and hence:
(3)
\begin{align} \quad \mathrm{int} (A) \cap \mathrm{int} (B) \subseteq \mathrm{int} (A \cap B) \end{align}
• Now let $x \in \mathrm{int} (A \cap B)$. Then there exists a $U \in \tau$ such that $x \in U \subseteq A \cap B$. But if $U \subseteq A \cap B$ then $U \subseteq A$ and $U \subseteq B$. So $x \in U \subseteq A$ and $x \in U \subseteq B$
• Therefore $x \in \mathrm{int} (A)$ and $x \in \mathrm{int} (B)$, so $x \in \mathrm{int} (A) \cap \mathrm{int} (B)$ and hence:
(4)
\begin{align} \quad \mathrm{int} (A) \cap \mathrm{int} (B) \supseteq \mathrm{int} (A \cap B) \end{align}
• We therefore conclude that $\mathrm{int} (A) \cap \mathrm{int} (B) = \mathrm{int} (A \cap B)$. $\blacksquare$