Countable union of relatively compact sets

Question

Let $X$ be a topological space and $\mathcal K(X)$ be $\sigma$-algebra, generated by compacts of $X$. Prove that for any set $B \in \mathcal K(X)$ either $B$ or its complement can be represented as a countable union of relatively compact sets.

My attempt: Let $B\in K(X)$ be a proper subset of $X$. Since $B \in \mathcal K(X)$ either it is a countable union of compact sets or its complement is such. Since all compact sets are relatively compact, we're done.

So the whole question is if $X$ can be represented as a countable union of relatively compact sets. And I'm stuck on this. Could you help me?

Answer 1

By definition, $\mathcal{K}(X)$ is the smallest $\sigma$-algebra which contains the compacts.

Let $\mathcal{A}$ denote the collection of all sets $A \subset X$ such that either $A$ or its complement $A^C$ is a countable union of relatively compact sets. Obviously $\mathcal{A}$ contains the compacts so, if $\mathcal{A}$ is also a $\sigma$-algebra, then $\mathcal{K}(X) \subset \mathcal{A}$ must hold.

• It is obvious that $\mathcal{A}$ is closed under complementation.
• Suppose $A_1,A_2,A_3\ldots \in \mathcal{A}$. If one of their complements, say $A_1^C$, can be expressed as a countable union of relatively compact sets, then so too can $\left( \bigcup_{i=1}^\infty A_i \right)^C = \bigcap_{i=1}^\infty A_i^C$ (why?). Otherwise, one must have that $A_i$ is a countable union of relatively compact sets for each $i$. In this case $\bigcup_{i=1}^\infty A_i$ can also be expressed in this form (why?).