Compactness

Template:Mergeto A space is compact if every open cover admits a finite subcover. This can be formalized as follows:

A topological space $X$ is compact , if for every family ${(U_i)}_{i\in I}$ of open sets with $X=\bigcup _{i\in I}{U}_i$ there exists a finite set $F\subseteq I$ such that $X=\bigcup _{i\in F}{U}_i$.

Compactness can also be expressed by one of the following equivalent characterisations:

• Every filter on $X$ containing a filter basis of closed sets has a nonempty intersection.
• Every ultrafilter on $X$ converges.

Theorem Every closed subset of a compact set is compact.
Proof: Let $X$ be a compact, and let $K$ be a closed subset of $X$. Consider any open cover C of $K$. Then consider the open cover C∪{X$\setminus$K} of X. This has a finite subcover. X$\setminus$K must be within that subcover because it would be required to cover an element not within $K$. The other sets within that subcover would thus cover K. This is a finite subcover of $K$.

Theorem Every compact subset of a Hausdorff space is closed.
Proof: Let $K$ be compact. If the complement $K^c$ is empty, then $K$ is the same as the space; thus closed. Suppose not; that is, there is a point $x\in K^c$. Then for each $y\in K$, by the Hausdorff separation axiom we can find $U_y$ and $V_y$ disjoin, open and such that $x\in U_y$ and $y\in V_y$. Since $K$ is compact and the collection $\{V_y;y\in K\}$ covers $K$, we can find a finite number of points $y_1,y_2,...y_n$ in $K$ such that:

$K\subseteq V_{y_1}\cup V_{y_2}\cup \dots \cup V_{y_n}$

It then follows that:

$x\in U_{y_1}\cap U_{y_2}\cap \dots \cap U_{y_n}\subseteq K^c$

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Theorem Every compact set in a metric space is bounded.
Proof: Consider the open balls centered upon a common point, with any radius. This can cover any set, because all points in the set are some distance away from that point. Any finite subcover of this cover must be bounded, because it will bound it within the largest open ball within that subcover. Therefore, any set covered by this subcover must also be bounded.

Heine-Borel Theorem For any interval [a,b], for any open cover C of that interval, there exists a finite subcover of C.
Proof: Define c=sup{x|[a,x] is covered by a finite number of sets within C}. This set is non-empty because a is within the set. Suppose that c<b. Then there is a finite cover of sets within C for [a,c]. c is within a set A within the cover C. Thus, there exists a ε>0 such that (c-ε, c+ε)⊆A. Then c+$\frac{ϵ}{2}$ is also within {x|[a,x] is covered by a finite number of sets within C}, contradicting the definition of c. Thus, c≥b. Therefore, there is a finite subcover of C.

Theorem If f:X→Y is continuous, and A is a compact set, then the image of A, f(A), is compact.
Proof: Let C be any open cover of f(A). Consider the inverses {f^-1(B)} where B∈C. These inverses are open because f is continuous. This covers A, and thus there is a finite subcover of A, {B_i}. Then the images {f(B_i)} is a finite subcover of f(A).

Theorem If a set is compact and Hausdorff, then it is normal.
Proof: Let X be compact and Hausdorff. Consider two closed subsets A and B which are themselves compact by one of the theorems above. For each element a∈A, there exists for each element b∈B, two disjoint sets O_{a, b, 1} and O_{a, b, 2} such that a∈O_{a, b, 1} and y∈O_{a, b, 2}. The union of all such O_{a, b, 2} for a fixed a is a cover for B, and thus has a finite subcover, and denote it O_{a, 2}, and its union O'_{a, 2}. Let B_a={b|O_{a, b, 2}∈O_{a, 2}}, and let O_{a, 1} denote the intersection of the set {O_{a, b, 1}|b∈B_a} which is an open set. The union of all such O_{a, 1} covers A, and therefore has a finite subcover, and denote its union to be U. Let A denote the set A' denote all elements of A such that O_{a, 1} is within the cover. Take the intersection of {O'_{a, 2}|a∈A'}, which is an open set, and denote it to be V. Then U is an open superset of A, V is an open superset of B, and they are disjoint. Thus, X is normal.