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Given a space $X$ and $C \subset X$, we say that $C$ is strongly discrete if there exists a disjoint family $\{U_x: x\in C\} $ of open sets in $X$ such that $x\in U_x$ for all $x\in C$. The question is this:

How to show any convergent sequence is strongly discrete in Hausdorff space?

Thanks ahead.

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We have to take $C$ which consists only of the terms of the sequence (not the limit $x_{\infty}$, otherwise it won't work). We can assume that $x_i\neq x_j$ if $i\neq j$. For $i,j\in\Bbb N\cup_{\infty}$, $i\neq j$, we can find disjoint open sets $O_{i,j}$ and $O'_{i,j}$ such that $x_i\in O_{i,j}$, $x_j\in O'_{i,j}$.

For all $i\in\Bbb N$, we can find $N_i\in\Bbb N$ such that $x_j\notin O_{i,\infty}$ if $j\geq N_i$. We can assume $N_i>i$. Then take $U_i:=\bigcap_{k=1}^{N_i}O_{ik}$: it's an open set. These ones are disjoint.

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    $\begingroup$ So: the correct statement seems to be: let $x_n \to x_{\infty}$ be a convergent sequence in a Hausdorff space. Then $C = \{x_n : n \in \mathbb{N}\} \setminus \{x_\infty\}$ is strongly discrete. An easy counterexample to the statement in the question is $x_0 = 0$ and $x_n = 1/n$ for $n \geq 1$. $\endgroup$
    – commenter
    Oct 28, 2012 at 9:37

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