Let $(X, \tau)$ be a topological space. Let $A:=\{X -U: U \in \tau\}$ (i.e. the set containing all closed subsets) and $K$ is a set containing all compact subspaces. Show:
i) $\sigma(\tau)=\sigma(A)$
My ideas: Let $B \in \tau$, it follows $B$ is open $\Rightarrow$ $B^{c}$ is closed and by definition $B^{c} \in A \Rightarrow \sigma(A)$ is a $\sigma-$Algebra, however, thus $^{c}-$Stable and so $(B^{c})^{c}=B \in \sigma(\tau)$. I still believe that I have not even shown the inclusion really here. All I showed was that:
$\tau \subseteq \sigma(A)$
If I can show that $ A\subseteq \sigma(\tau)$, can I immediately say that I have proved $\sigma(\tau)=\sigma(A)$?
ii) If $X$ is $\sigma-$compact (i.e. Countable union of compact spaces) Hausdorff-Space, then $\sigma(\tau)=\sigma(K)$
My ideas: Not at all sure on this one, but I'll give it a go. Since $X$ is compact and a Hausdorff space. Any compact subspace will be closed. Let $(A_{n})_{n}$ be any collection of compact subspaces in $X$ $\Rightarrow (A_{n}^{c})_{n}$ is a collection of open sets, $A_{n}^{c} \in \tau, \forall n \in \mathbb N$, and thus like in i), we prove that $\tau \subseteq \sigma(K)$
Am I on the right track. It does not feel like it. Any help would be strongly appreciated.
Your idea for part i) is pretty much correct. Since $\tau\subseteq \sigma(A)$, it follows that $\sigma(\tau)\subseteq \sigma(A)$, because $\sigma(\tau)$ is the smallest $\sigma$-algebra containing $\tau$.
I am not sure about your ideas for part ii) though. Since the space is Hausdorff, we know as you mentioned that every compact set is closed and therefore $K\subseteq A$. It follows that $\sigma(K)\subseteq\sigma(A)=\sigma(\tau)$. Now since $X$ is $\sigma$-compact, we can find a sequence of compact sets $C_n$ so that $X=\cup C_n$. Then if $S\in A$ is a closed set, we can write $S=\cup_n(S\cap C_n)$. Moreover every set $S\cap C_n$ is a closed subset of $C_n$, hence compact. It follows that $S\in\sigma(K)$, and therefore $A\subseteq\sigma(K)$. As in part i), this implies that $\sigma(A)\subseteq\sigma(K)$ and we're done.