Yesterday I solved for pseudo compact sets which are closed and bounded, and with the help of stack exchange I learnt that any subspace is Lindelöf.My problem is only $\sigma$-compact set that has not been solved.
My question is what kind of subspaces of $\mathbb{R}^n$ would be $\sigma$-compact.
I have solve that $\sigma$-compactness is not equivalent to bounded and closed, how could it would be?
Some clues will be grateful.Thanks!
Note that a $\sigma$-compact subspace $A\subset(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$ is a countable union of compact sets. Since a compact set in $(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$ is closed, $A$ is an $F_{\sigma}$. Conversely, suppose $A$ is an $F_{\sigma}$ set in $(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$, namely, $A$ is a countable union of closed sets. To verify that $A$ is $\sigma$-compact it suffices to show that all the closed sets in $(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$ are $\sigma$-compact, because the countable union of countable sets is still countable. Assume that $B$ is an arbitrary closed set in $(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$. Define $Q_k$ to be the closed square $[-k,k]^n$ in $\mathbb{R}^n$. Observe that for a fixed $k\in\mathbb{N}$, $Q_k$ is closed and bounded. From Heine-Borel theorem, it follows that $Q_k$ is also compact. Moreover, $Q_k\cap B$ is compact, because it is closed in a compact subspace $Q_k$. In light of $$B=\bigcup_{k\in\mathbb{N}}Q_k\cap B,$$ we have that $B$ is $\sigma$-compact, which completes the proof.