Borel sigma-algebra for compacts

Question

How can we show that the borel sigma-algebra on the interval (0, 1] is the smallest sigma-algebra that contains compact sets ?

Answer 1

Since $(0,1]$ is a subset of $\Bbb R$, we know compactness is equivalent to being closed and bounded. Since every subset of $(0,1]$ is bounded, this implies subsets of $(0,1]$ are compact if and only if they are closed.

Let $\mathcal{C}$ be the collection of closed (i.e., compact) subsets of $(0,1]$. Suppose by contradiction that there is a smaller $\sigma$-algebra $\mathcal{B}'$ than the Borel $\sigma$-algebra which contains $\mathcal{C}$.

Since $\mathcal{B}'$ contains $\mathcal{C}$ and it is also closed under set complements, $\mathcal{B}'$ contains all open sets in $(0,1]$. But then this is a $\sigma$-algebra that's smaller than the Borel $\sigma$-algebra which contains the open sets, and this is a clear contradiction. So $\mathcal{B}'$ doesn't exist, and thus the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing the closed subsets of $(0,1]$, and thus the compact subsets of $(0,1]$.

Answer 2

The $\sigma$-algebra generated by the compacts contains every open subinterval, so it contains every open subset. Therefore, it contains the Borel sets.