If $A$ is a Hausdorff space such that $A = \bigcup\limits_{i = 1}^\infty {{K_i}}$ where $K_i$ are its compact subsets, is $A$ a paracompact space? If not, what additional conditions should we add? (e.g. locally compact)

• Don't know off-hand, but adding regularity would suffice, since a countable union of compact sets is Lindelof, and regular+Lindelof implies paracompact. – Arturo Magidin Dec 8 '11 at 17:14

Proof: Suppose that $X$ is a paracompact Hausdorff space, $F\subseteq X$ is closed, and $x\in X\setminus F$. Since $X$ is Hausdorff, each $y\in F$ has an open nbhd $V_y$ such that $x\notin\operatorname{cl}V_y$, and the sets $V_y$ ($y\in F$) together with $X\setminus F$ are an open cover of $X$. Let $\mathscr{W}\;$ be a locally finite open refinement of this cover, and let $U=\bigcup\{W\in\mathscr{W}:W\cap F\ne\varnothing\}$; $U$ is an open set containing $F$. Locally finite collections are closure-preserving, so $\operatorname{cl}U=\bigcup\{\operatorname{cl}W\in\mathscr{W}:W\cap F\ne\varnothing\}$. Suppose that $W\in\mathscr{W}$ meets $F$. Then $W\subseteq V_y$ for some $y\in F$, so $\operatorname{cl}W\subseteq\operatorname{cl}V_y$, and hence $x\notin\operatorname{cl}W$. It follows that $x\notin\operatorname{cl}U$ and hence that $X\setminus\operatorname{cl}U$ and $U$ are disjoint open nbhds of $x$ and $F$. $\dashv$
A space that is the union of countably many compact subsets is said to be $\sigma$-compact. Clearly every countable Hausdorff space is $\sigma$-compact, and there are countable Hausdorff spaces that are not regular and hence not paracompact, e.g., the irrational slope topology, originally described by R.H. Bing in this paper.
A Hausdorff $\sigma$-compact space is paracompact iff it is regular; the necessity follows from the regularity of paracompact Hausdorff spaces, and as Arturo pointed out, regularity is sufficient because a $\sigma$-compact space is Lindelöf, and a regular Lindelöf space is paracompact.