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We're concerned only with a normed vector space here.

The book "Introduction to Functional Analysis" by Kreyszig says that if we take the closure $\overline{A}$ of set $A$, then for every $x\in \overline{A}$, there exists a sequence $(x_j)$ converging to it.

I've never come across this fact before. On thinking about it, I feel this would imply that every point in $\overline{A}$ is a limit point of $\overline{A}$, which is not a necessary condition. Another possibility is that in $(x_j)$, all the terms $x_n$ for $n\geq N$ are $x$.

Could someone kindly shed light on the fact stated in the book? How else could it be possible that in a closed set, every point is the limit of a sequence, such that both the point and sequence are entirely within the closed set? Also, why would this not be true in an open set? I understand that the limit points of all cauchy sequences need not be inside an open set. However, every point can still be the limit point of some cauchy sequence inside the open set.

Thanks in advance!

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I think what you missed is this. Every element $x$ in $\bar A$ is the limit of a sequence $x_n$ where all the elements of $x_n$ are in $A$.

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  • $\begingroup$ Oh, I actually had that in mind. So unless $x$ is a limit point of $A$, wouldn't this be possible only if all $x_n\in (x_j)$, where $n\geq N$, are equal to $x$? $\endgroup$ – fierydemon Nov 7 '13 at 4:24
  • $\begingroup$ If $x\in A$, then the sequence you mention will definitely work. It won't necessarily be correct that this is the only possible way. But for some choices of $A$ it will be. $\endgroup$ – Stephen Montgomery-Smith Nov 7 '13 at 13:20

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