I wonder to know what sets different types of compact subspaces in $\mathbb{R}^n$ with the standard topology are.
I already solve for compact sets, countably compact sets, sequentially compact sets, limits point compact sets, which are closed and bounded.
But I fail for the Lindelöf subspaces.
Some clues will be grateful.Thanks!
For an arbitrary subset $A\subset(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$, we intend to prove that $A$ is Lindelöf. we need two lemmas.
A $\mathrm{C}_2$ space is Lindelöf.
Second-countability is hereditary.
A brief proof for the first lemma is as follows. Let $(X,\mathcal{T})$ be a $\mathcal{T}_2$ space and let $\mathcal{B}={B_n\colon n\in \mathbb{N}}$ be a countable basis of $(X,\mathcal{T})$. Consider an open cover $\mathcal{U}=\{U_i\colon i\in I\}$. Define $J\subset\mathbb{N}$ by $n\in J$ if and only if there exists $U\in\mathcal{U}$ such that $B_n\subset U$. As a subset of $\mathbb{N}$, $J$ is a countable set and for every $n\in J$ the set $\mathcal{U}_n=\{U\in\mathcal{U}\colon B_n\subset U\}$ is non-empty, by definition of $J$. From the axiom of choices, it follows that there is a function $f:J\to\cup_n\mathcal{U}_n$ such that $f(n)\in\mathcal{U}_n$ for all $n\in J$. For any $x\in X$, there exists $U_x\in\mathcal{U}$ containing $x$, as $\mathcal{U}$ is a cover of $X$. By definition of basis, we have $x\in B_{n_x}\subset U_x$, for some $n_x\in\mathbb{N}$. Thus, $n_x\in J$ and $f(n_x)\in\mathcal{U}_{n_x}$. Consequently, $x\in B_{n_x}\subset f(n_x)\in\mathcal{U}'$, which implies that $\mathcal{U}'=\{f(n)\colon n\in J\}$ is a countable subcover of $X$ indexed by $J$.
The second lemma is obvious by taking the subspace topology.
Now we prove the conclusion. For an arbitrary subset $A\subset(\mathbb{R}^n,\mathcal{T}_{\mathbb{R}^n})$, since $\mathrm{C}_2$ is hereditary, $A$ also satisfies $\mathrm{C}_2$ axiom. In light of the second lemma, $A$ is Lindelöf.