Heine Borel Theorem

This is the simplest form of the HBT, specialized to the unit interval of the real line.

Let $O$ be an collection of open intervals which cover the unit interval $[0,1]$ . Then there is a finite subcover.

Let $G$ denote the collection of all $x$ such that $x \le 1$ and $[0,x]$ is has a finite subcover in $O$ . Note that $0 \in G$ . Hence $g = \sup G$ exists and is not larger than $1$ .

first claim: $g \in G$ .

Assume $g \not \in G$ . Since $O$ is an open cover of $[0,1]$ there is an open interval $(a,b) \in O$ such that $x\in (a,b)$ . Since $[0,(a+x)/2]$ has a finite subcover in $O$ it follows that adding $(a,b)$ to this subcover is a finite subcover of $[0,g]$ . Hence $g \in G$ , which is a contradiction to the assumption.

second claim: $g = 1$ .

Assume $g < 1$ . Since $[0,g]$ has a finite subcover, it has an open interval $(a,b)$ such that $g \in (a,b)$ . So $(g+b)/2 \in G$ , and so $g < (g+b)/2 \le \sup G$ . This is a contradiction since $\sup G = g$ .

Hence the conclusion is validated.

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