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Let $F \subset \mathbb{R}^n$. A point $x \in \bar F$ iff there exists a sequence $\{x_k\} \subset F$ such that $x_k \to x$.

How can I mathematically show this statement?

I'll define the closure of the set as follows:

Given a set $F \subset \mathbb{R}^n$ the closure of the set $\bar F := \{y \in \mathbb{R}^n \space \text{for all} \space r > 0, \text{it holds that} \space B(y,r) \cap F \neq \emptyset \}$.

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  • $\begingroup$ @FamousBlueRaincoat - thanks for that, I've edited my original post to include a definition. $\endgroup$
    – yaweh
    Feb 7 '15 at 3:00
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If $x \in \bar{F}$, then choose $x_n \in B(x,{1 \over n}) \cap F $. Then $x_n \to x$.

Now suppose $x_ n \to x$, with $x_n \in F$. Let $\epsilon>0$, then there is some $n$ such that $x_n \in B(x,\epsilon) \cap F$. Since $\epsilon>0$ was arbitrary, we see that $x \in \bar{F}$.

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  • $\begingroup$ Clear and concise! Was having difficulty putting my thoughts into words. Thank you :) $\endgroup$
    – yaweh
    Feb 7 '15 at 3:39

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