Compactness Review

Compactness Review

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

(1)
\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.
• 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:
(2)
\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 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 $Y$.
• 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 $Y$. 