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Is it true that if $C$ is a closed set, and $x \in {C}$ then there exist a sequence $\{x_i\}$ which tends to $x$?

I'm not sure about the correctness of this statement, and whether it is true for every point in the set $C$ or for certain points only (boundary points?). And of course - why is it true? under which conditions?

Elad Maimoni
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    Constant sequence ${x_i } = {x }$ ? – charmd Apr 27 '16 at 11:20
  • The sequence $x_n=x$ for all $n$ tends to $x$. But if you mean a sequence with $x_n\ne x$ for all $n$, then the answer is no. For example, the set ${1}$ contains all its limit points, so it is closed. – almagest Apr 27 '16 at 11:21
  • Oh, do you require that the points $x_n$ are in $C$? – almagest Apr 27 '16 at 11:22
  • OK I'll try to be a bit more clear about the context:

    http://math.stackexchange.com/questions/202378/is-closure-of-convex-subset-of-x-is-again-a-convex-subset-of-x/202380#202380

    I'm looking at the first answer and can't figure out the 'exist' part. feel free to edit my question so my intention will be more clear.

     – Elad Maimoni Apr 27 '16 at 11:27
  • @TheKeyBoardHero So it seems you do want the $x_i$ in $C$, but you don't care if they are all equal to $x$. So on that basis, yes it is true that there exists a sequence $x_n$ which tends to $x$. – almagest Apr 27 '16 at 11:31
  • @TheKeyBoardHero Use the definition of the closure (see https://en.wikipedia.org/wiki/Closure_(topology)). An element is in the closure of $C$ iff there is a sequence in $C^{\mathbb{N}}$ which converges towards your element – charmd Apr 27 '16 at 11:34
  • @almagest I don't quite agree. In the context I refered to, $x_i \in X$ but the limit $x$ belongs in $\bar{X}$. – Elad Maimoni Apr 27 '16 at 11:37
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    @TheKeyBoardHero Yes, but if $X$ is closed (which is the case for your set in this question), then the closure of $X$ is $X$ itself – charmd Apr 27 '16 at 11:38

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