I think it is correct, but I would like another pair of eyes to verify.

Definition. An open neighborhood assignment is a function $f:X\to \tau$ such that $x\in f(x)$.

Definition. A space is said to be D, or a D-space, if for every open neighborhood assignment $f:X\to \tau$ there exists a closed discrete set such that $f[D]$ is a cover of $X$.

Theorem. Every metric space is a D-space

Proof. Let $f:X\to\tau$ be an open neighborhood assignment in a metric space. Then define the function $\textit{d}:X \to \mathbb{N}$ by $\textit{d}(x) = \min\lbrace n\in\mathbb{N} | B_{1/n}(x)\subset f(x)\rbrace$. Since this is possible, we will actually just replace our open neighborhood assignment with the smaller assignment given by $\lbrace B_{1 / \mathit{d}(x)}(x)\in \tau | x\in X \rbrace$. If it is possible to show that there is a one such closed discrete set using this smaller assignment, then it is then possible to use the original assignment. For all $n\in\mathbb{N}$, define $L_n = \lbrace x\in X|\textit{d}(x) = n\rbrace$. Enumerate $X=\lbrace x_\alpha | \alpha<\lambda\rbrace$ such that if $\alpha<\beta$ then either $x_\alpha,x_\beta\in L_i$ for some $i$ or $x_\alpha\in L_i$ and $x_\beta\in L_j$ for some $i<j$. Without loss of generality, suppose that $L_1$ is non-empty. Define $D_1 = \lbrace x_1 \rbrace$. If $x_\alpha$ is not in $f[\cup_{\beta < \alpha}D_{\beta}]$, then let $D_\alpha = \cup_{\beta < \alpha}D_{\beta}\cup \lbrace x_\alpha \rbrace$. If $x_\alpha$ is in $f[\cup_{\beta < \alpha}D_{\beta}]$ then we let $D_\alpha = \cup_{\beta < \alpha}D_{\beta}$. Let $D=\bigcup_{\alpha\in\lambda}D_\alpha$.

Claim. D is discrete.

Argument. (no one is in another person's bubble) Let $x_\alpha,x_\beta\in D$. Since $X$ is well ordered, let $\alpha<\beta$. Then $x_\beta \notin f(x_\alpha)$ by construction. Hence, $\rho(x_\alpha,x_\beta)\geq 1/ \textit{d}(x_\alpha)$. But also, since $\alpha<\beta$, $1/ \textit{d}(x_\alpha)\geq 1/ \textit{d}(x_\beta)$. Thus we have $\rho(x_\alpha,x_\beta)\geq 1/ \textit{d}(x_\alpha)\geq 1/ \textit{d}(x_\beta)\Rightarrow x_\alpha\notin f(x_\beta)$.\

Claim. D has no accumulation points

Argument. If $D$ were to have an accumulation point, it surely won't be an element of $D$ by our previous result. Hence let $x_\alpha\in X\setminus D$. Since $x_\alpha$ was not chosen to be in $D$, it follows that there exists $\beta<\alpha$ such that $x_\alpha\in f(x_\beta)$. Let $\epsilon>0$ such that $B_\epsilon(x_\alpha)\subset f(x_\beta)$. By our previous result, there is no $y\in D$ such that $y\in f(x_\beta)$. Hence, no $y\in D$ such that $y\in B_\epsilon(x_\alpha)$.

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    $\begingroup$ You mean to take $B_{\color{red}{1/}d(x)}(x)$ as smaller assignment, donÄt you? $\endgroup$ Dec 15 '14 at 7:19
  • $\begingroup$ Yes, thank you. $\endgroup$
    – Prototank
    Dec 15 '14 at 7:21
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    $\begingroup$ It's fine, apart from a typo in the proof that $D$ is discrete: You want $\color{red}{1/}d(x_\alpha)\ge 1/d(x_\beta)$. (Also, you don't need to split the induction into two cases: your argument for the limit case works for both.) $\endgroup$ Dec 15 '14 at 8:07

Your proof is fine. (It's just that by personal taste I tend to try and use only a single, common definition for successor and limit ordinals)


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