Compactness Review

# Compactness Review

We will now review some of the recent material regarding compactness in topological spaces.

- Recall from the
**Covers of Sets in a Topological Space**page that if $X$ is a topological space then an a**Cover**of $X$ is a collection of sets $\mathcal F$ such that:

\begin{align} \quad X \subseteq \bigcup_{U \in \mathcal F} U \end{align}

- If the sets in $\mathcal F$ are all open (or closed) and satisfy the inclusion above then $\mathcal F$ is called an
**Open Cover**(or**Closed Cover**) of $X$.

- On the
**Compactness of Sets in a Topological Space**page we said that a set $A \subseteq X$ is said to be**Compact**in $X$ if every open cover of $A$ has a finite subcover. We saw that the set of natural numbers $\mathbb{N}$ is NOT compact in $\mathbb{R}$ with the usual topology.

- On the
**Compactness of Finite Sets in a Topological Space**we saw a very nice theorem regarding the compactness of finite sets. We saw that if $A = \{ x_1, x_2, ..., x_n \}$ is a finite subset of a topological space $A$ is compact in $X$. This is because if we are given an open cover $\mathcal F$ of $A$ then all we need to do is select a set in the open cover associated to each point $x_i \in A$, and so, we will always be able to construct a finite subcover of size less than or equal to $n$.

- As a result, every subset of a finite topological space is compact.

- On the
**Finite Intersection Property Criterion for Compactness in a Topological Space**page we looked at a nice result with regards to compactness of a topological space.

- First, if $\mathcal F$ is any collection of sets then we said that $\mathcal F$ has the
**Finite Intersection Property**if every finite collection of sets $\{ F_1, F_2, ..., F_n \} \subseteq \mathcal F$ we have that:

\begin{align} \quad \bigcap_{i=1}^{n} F_i \neq \emptyset \end{align}

- In other words, the intersection of all sets in a finite subset of $\mathcal F$ is nonempty.

- We then proved that a topological space $X$ is compact if and only if for every collection of closed sets $\mathcal F$ of $X$ that have the finite intersection property then the intersection of all sets in $\mathcal F$ is nonempty, i.e. $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$.

- On the
**Preservation of Compactness under Continuous Maps**we saw that if $f : X \to Y$ is a continuous map and $X$ is a compact topological space then the range $f(X)$ is also a compact space.

- On the
**Closed Sets in Compact Topological Spaces**we saw that if $X$ is a compact topological space, $A \subseteq X$, and $A$ is closed in $X$ then $A$ is also compact in $X$.

- On the
**Compact Sets in Hausdorff Topological Spaces**page we proved that if $X$ is a Hausdorff topological space (i.e., for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$) then every compact set $A \subseteq X$ is closed.

- We then began to see how useful the concept of a compact topological space was. On the
**Homeomorphisms Between Compact and Hausdorff Spaces**page we saw that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous bijective function, then $f$ is necessarily a homeomorphism between $X$ and $Y$.

- Furthermore, on the
**Quotients from Equivalence Relations defined by Functions from Compact to Hausdorff Spaces**we first proved that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous map then $f$ is also a closed map (i.e., the image of any closed set in $X$ is closed in $Y$).

- We then proved a nice result. We proved that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f : X \to Y$ is continuous and surjective then for the equivalence relation $\sim_f$ defined for all $x_1, x_2 \in X$ by $x_1 \sim_f x_2$ if and only if $f(x_1) = f(x_2)$, we have that the quotient space $X /\: \sim_f$ is homeomorphic to $Y$.