Closure, Boundary, and Hausdorff Topological Spaces Review

# Closure, Boundary, and Hausdorff Topological Spaces Review

We will now take some time to review the concepts of the closure and boundary of a topological space and the special type of topological spaces known as Hausdorff spaces.

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

• Recall from The Closure of a Set in a Topological Space that if $A \subseteq X$ then the Closure of $A$ is the smallest closed set containing $A$ and is denoted $\bar{A}$. Since $\bar{A}$ contains $A$ we have that:
(1)
\begin{align} \quad A \subseteq \bar{A} \end{align}
• For example, the closure of $A = (0, 1)$ in the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$ is $\bar{A} = [0, 1]$.
(2)
\begin{align} \quad \bar{A} = A \cup A' \end{align}
• On The Closure of Closed Sets in a Topological Space page we saw that a set $A \subseteq X$ is closed if and only if $\bar{A} = A$. Consequentially, $\bar{\emptyset} = \emptyset$ and $\bar{X} = X$. We also noted that if $A$ is a clopen set then $\bar{A} = A$.
(3)
\begin{align} \quad \bar{A} \subseteq \bar{B} \end{align}
• Furthermore, if $A, B \subseteq X$ then the union of the closure of $A$ with the closure of $B$ is equal to the closure of the union of $A$ with $B$:
(4)
\begin{align} \quad \bar{A} \cup \bar{B} = \overline{A \cup B} \end{align}
• Additionally, the intersection of the closure of $A$ with the closure of $B$ is a superset of the closure of the intersection of $A$ with $B$:
(5)
\begin{align} \quad \bar{A} \cap \bar{B} \supseteq \overline{A \cap B} \end{align}
The Interior of $A$, $\mathrm{int} (A)$ The Closure of $A$, $\bar{A}$
The interior of $A$ is the LARGEST OPEN set CONTAINED in $A$ The closure of $A$ is the SMALLEST CLOSED set CONTAINING $A$
$\mathrm{int} (A) \subseteq A$ $A \subseteq \bar{A}$
$A$ is OPEN if and only if $A = \mathrm{int} (A)$ $A$ is CLOSED if and only if $A = \bar{A}$
If $A \subseteq B$ then $\mathrm{int} (A) \subseteq \mathrm{int} (B)$ If $A \subseteq B$ then $\bar{A} \subseteq \bar{B}$
$\mathrm{int} (A) \cup \mathrm{int} (B) \subseteq \mathrm{int} (A \cup B)$ $\bar{A} \cup \bar{B} = \overline{A \cup B}$
$\mathrm{int} (A) \cap \mathrm{int} (B) = \mathrm{int} (A \cap B)$ $\bar{A} \cap \bar{B} \supseteq \overline{A \cap B}$
• On the The Boundary of a Set in a Topological Space page we defined a point $x \in X$ to be a Boundary Point of $A$ if $x \in \bar{A} \setminus \mathrm{int} (A)$. The set of all boundary points of $A$ is called the Boundary of $A$ and is given by:
(6)
\begin{align} \quad \partial A = \bar{A} \setminus \mathrm{int} (A) \end{align}
• Soon after on The Boundary of Any Set is Closed in a Topological Space page we noted that the boundary of any set is always closed, i.e., $\partial A$ is always closed. We proved this by showing the complement $(\partial A)^c = X \setminus \partial A$ was open by writing $X \setminus \partial A = (A \setminus \partial A) \cup (A^c \setminus \partial A)$ and showing that $X \setminus \partial A$ was therefore the union of two open sets implying $X \setminus \partial A$ was open and $\partial A$ is closed.
• On the The Boundary of Clopen Sets in a Topological Space page we looked at a very important theorem which told us that a set $A$ is clopen if and only if $\partial A = \emptyset$. We proved this since if $A$ is clopen (open and closed) then smallest closed set containing $A$ is $A$ so $\bar{A} = A$, and the largest open set contained in $A$ is $A$ so $\mathrm{int} (A) = A$. Therefore $\partial A = \bar{A} \setminus \mathrm{int} (A) = A \setminus A = \emptyset$.
• Consequentially, since $\emptyset$ and $X$ are both clopen sets we saw that:
(7)
\begin{align} \quad \partial \emptyset = \emptyset \quad \partial X = \emptyset \end{align}
(8)
\begin{align} \quad \partial A \cup \partial B \supseteq \partial (A \cup B) \end{align}
• Lastly, on the Hausdorff Topological Spaces page we said that $(X, \tau)$ is a Hausdorff Topological Space if for every distinct $x, y \in X$ there exists a $U, V \in \tau$ with $x \in U$ and $y \in V$ such that $U \cap V = \emptyset$ - i.e., any distinct pair of points in the topological space can be separated with a pair of disjoint open neighbourhoods. We noted that the set of real numbers with the usual topology is a Hausdorff space because no matter what points $x, y \in \mathbb{R}$ we choose there always exists open intervals around $x$ and $y$ that do not intersect