# A question on a sequence

Let $X$ be Hausdorff. If $S=\{x_n: n\in \mathbb N \}\subset X$, then does there exist a subsequence $D$ of the sequence $S$ such that $D$ is a discrete subset of $X$?

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Have you tried working this out for some concrete examples of $X$ and $S$? –  Brad Jan 17 '13 at 12:40
I think you mean "$D$ is a discrete subspace of $X$. (That is, every point of $D$ is open in the subspace topology of $D$). This may be of interest then. –  David Mitra Jan 17 '13 at 13:06
@DavidMitra: You are RIGHT. –  Paul Jan 17 '13 at 13:11
–  Martin Jan 17 '13 at 13:20
So use the fact that if $A_1\subset A_2\subset X$, then the relative topology induced on $A_1$ by the relative topology of $A_2$ in $X$ is precisely the relative topology of $A_1$ in $X$, and fact that every infinite Hausdorff space contains a countably infinite discrete subspace. The latter fact is proven in this post. –  David Mitra Jan 17 '13 at 13:20

If $x \in X$ has a finite open nbhd, then $\{x\}$ is open (i.e., $x$ is an isolated point). (Why?) Let $I$ be the set of isolated points of $X$; clearly $I$ is discrete, so if $I$ is infinite, just pick any countably infinite subset of it. That's the easy case; you have to work harder when $I$ is finite.
In that case let $Y = X \setminus I$. Note that if $y \in Y$, every open nbhd $V$ of $y$ is infinite. Now build the desired discrete set recursively. Start by choosing distinct points $y_0,y_1 \in Y$, and let $W_0,V_1$ be disjoint open sets with $y_0 \in W_0$ and $y_1 \in V_1$. (Here there's a reason for using different letters: the $V_n$'s are needed temporarily for the construction, but it's the $W_n$'s that we really want.) $V_1 \cap Y$ is infinite (why?), so we can pick a point $y_2 \in V_1 \setminus \{y_1\}$. Now let $W_1$ and $V_2$ be disjoint open subsets of $V_1$ such that $y_1 \in W_1$ and $y_2 \in V_2$. Now $V_2 \cap Y$ is infinite, so we can pick a point $y_3 \in V_2 \setminus \{y_2\}$ and let $W_2$ and $V_3$ be disjoint open subsets of $V_2$ such that $y_2 \in W_2$ and $y_3 \in V_3$. Continue in this fashion to get points $y_n$ and open sets $W_n$ for each $n \in \omega$. Clearly $y_n \in W_n$ for each $n$, and with a little thought you should be able to see that if $m \ne n$, $y_m \notin W_n$. (It's probably best to consider the cases $m<n$ and $m>n$ separately. It may also help to make a sketch of the first few steps of the construction.)