September 2000

Problem Statement: Use the open cover definition of compactness to show that any finite union of compact sets is compact.

Proof: Let C_1,C_2,\dots,C_n be compact sets in \mathbb{R}. We wish to show that C=C_1\cup C_2\cup\dots\cup C_n is a compact set.

Let T=\bigcup\limits_{\alpha}T_{\alpha} be an open cover of C. Then T is an open cover for C_j for every j\in[1,n]. Since each C_j is compact there is a finite subcover of T which covers C_j for each j\in[1,n]. So for each C_j we may construct T_{j}=T_{j_1}\cup T_{j_2}\cup\dots\cup T_{j_m}, a finite subset of T which covers C_j, for each C_j.

I claim that T_C=\bigcup\limits_{j=1}^{n}T_j is a finite subcover of C.

First note that T_C is a union of finitely many unions of open sets and so it is an open set. Now to verify that T_C covers C. Let x\in C, then x\in C_j for some C_j. Thus, by construction, x\in T_j since T_j is an open cover of C_j. Therefore, x\in T_C and we may conclude that T_C is a finite subcover of T.

Therefore, every open cover of C has a finite subcover and so C is compact.

\Box

Reflection: The key in this proof is that we had a finite union of compact sets. From there all we had to do was union up all of their finite subcovers. A finite union of a finite union is simply a finite union of open sets.

Advertisements
This entry was posted in Analysis, Compact, Topology. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s