# how to prove that sigma-compact space is D-space

In the paper "A survey of D-spaces" by Gary Gruenhage it is written that it is easily seen that $\sigma$-compact spaces are D-spaces.

Unfortunately, I don't know how to show it. The considered spaces are regular and $T_1$. Thank you for your help! Karel Pastor

(Added by Asaf) The definition of a D-space, taken from the abstract of the aforementioned paper:

A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x) \colon x \in D\}$ covers $X$.

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I think this works: pick an increasing family of compact sets $K_n$, which covers $X$. Because of compactness, a finite subfamily of $\{N(x);x\in K_0\}$ covers $K_0$; denote their "base points" by $x^0_0,\ldots,x^0_{n_0}$. Now, by compactness again, a finite subfamily of $\{N(x)\cap K_0^c;x\in K_1\setminus\bigcup_{i=0}^{n_0}N(x^0_i)\}$ covers $K_1\setminus\bigcup_{i=0}^{n_0}N(x^0_i)$; denote their base points by $x^1_0,\ldots,x^1_{n_1}$. Continuing in this way, we get a set $D=\bigcup_{i=0}^\infty\{x^i_{0},\ldots,x^i_{n_i}\}$, the neighborhoods of whose points cover $X$.
Now pick a point $x^i_j\in D$. Using Hausdorffness, we can separate $x^i_j$ from all of the points in $D\cap K_i$. Since we picked the points $x^k_l, k>i,$ away from $N(x^i_j)$, we can separate $x^i_j$ from the points $(D\cap K_i)^c$. Therefore $x^i_j$ is an isolated point in $D$. This proves the discreteness of $D$. Almost the same argument shows that $D$ is closed in $X$.
Here goes a simpler proof (in my opinion). As in the previous one, it doesn't need to be assumed that the space is regular, the desired follows just by assuming Hausdorff. Notice that, in spaces that are at least $T_1$, $A$ is a closed discrete subset iff $\{\{x\}: x \in A\}$ is locally finite. So, take the very same increasing sequence of compact subsets, say $\{K_n: n < \omega\}$, such that $X = \bigcup\limits_{n < \omega} K_n$. Let $\langle x\rangle_0$ be a finite subset of $K_0$ such that $N[\langle x\rangle_0]$ covers $K_0$ (where, of course, $N[\langle x \rangle_0]$ is a short for $N[\textrm{im}(\langle x \rangle_0)]$). Then $$\{N(x): x \in K_1 \setminus N[\langle x\rangle_0]\}$$ covers the compact set $K_1 \setminus N[\langle x\rangle_0]$ - and this is the point where we use Hausdorffness, to be sure that $K_1$ is a closed set, and therefore $K_1 \setminus N[\langle x\rangle_0]$ is a closed subset of a compact set -, and so there is a finite sequence $\langle x \rangle_1$ of elements of $K_1 \setminus N[\langle x\rangle_0]$ such that $N[\langle x \rangle_0\, ^\frown \langle x \rangle_1]$ covers $K_1$. Proceeding inductively, if $j > 1$ there is a finite sequence $\langle x \rangle_j$ of elements of $K_j \setminus N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_{j -1 }]$ such that $N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_j]$ covers $K_j$. Now, with all finite sequences $\langle x \rangle_n$ constructed, let $x$ be the infinite sequence $$x = \langle x \rangle_0\,\frown\langle x \rangle_1 \frown \ldots \frown \langle x \rangle_n \frown \ldots$$ and take $F = im(x)$. It is clear that $N[F] = X$. We can check that $F$ is closed and discrete at once: it suffices to verify the local finiteness of the family of singletons. Indeed: it $t \in X$, let $$m = \textrm{min}\{j: t \in K_j\}.$$ By construction, $V = N[\langle x \rangle_0\, ^\frown \langle x \rangle_1\,\frown \ldots \frown \langle x \rangle_m]$ is an open neighbourhood of $t$ which contains only finitely many elements of $F$.