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:

\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 $\bar{A} = [0, 1]$.

- On
**The Closure of a Set Equals the Union of the Set and its Accumulation Points**page we looked at an equivalent definition of the closure of a set to be the union of the set itself with the set of all its accumulation points, that is:

\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 $\bar{A} = A$.

- We then looked at some theorems on the
**Basic Theorems Regarding the Closure of Sets in a Topological Space**page. We noted that if $A \subseteq B$ then:

\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 $B$:

\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 $B$:

\begin{align} \quad \bar{A} \cap \bar{B} \supseteq \overline{A \cap B} \end{align}

- On the
**A Comparison of the Interior and Closure of a Set in a Topological Space**page, we compared the interior $\mathrm{int} (A)$ and closure $\bar{A}$ of a set $A$ which is summarized in the following table:

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:

\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:

\begin{align} \quad \partial \emptyset = \emptyset \quad \partial X = \emptyset \end{align}

- On the
**Basic Theorems Regarding the Boundary of a Set in a Topological Space**page we could only note one important relationship. If $A, B \subseteq X$ then the union of the boundary of $A$ with the boundary of $B$ is a superset of the boundary of the union of $A$ with $B$:

\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 $v \in V$ 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