Let $E_0$ be the interval $[0,1]$. Remove the segment $\left(\frac{1}{3},
\frac{2}{3}\right)$, and let $E_1$ be the union of the intervals $$\left[ 0,
\frac{1}{3}\right], \left[\frac{2}{3},1\right]$$ Remove the middle thirds of these
intervals and let $E_2$ be the union of those intervals. Continuing in
this way we obtain a sequence of compact sets $E_n$, such that(a) $E_1 \supset E_2 \supset …$
(b) $E_n$ is the union of $2^n$ intervals, each of length $3^{-n}$.
The set $$P=\bigcap_{n=1}^\infty E_n$$ is called the Cantor set. $P$
is clearly compact (…)
I don't see why it is clearly compact. If it was a finite intersection I would be able to apply the Corollary of 2.35 which states that intersection of two compacts is compact (it actually states something more general but that's beyond the point). But it is not an infinite intersection. Why is it then so clear that $P$ is compact?
Best Answer
The set $P$ is closed as the intersection of closed sets. It is also bounded. By Heine-Borel - it is compact.