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A metric space $X$ is closed if every convergent sequence $(x_n)$ in $X$ converges in the space. A space is closed if it contains all its boundary points, so then I thought one could show a space is closed by showing that an arbitrary point in the space is a boundary point. This could be done by taking an arbitrary point $x \in X$, and showing that for every $\epsilon > 0 $ one can find a point $y\notin X $, such that $d(x, y) < \epsilon$.

I understood this is an invalid way to show a space is closed, since with this method I managed to show that $c_{00}$ (the space of sequences thats eventually only zeroes) is closed. But $c_{00}$ is not closed, so my proof method above must clearly be wrong.

Where is my proof method wrong?

landigio
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  • Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]\subseteq \mathbb{R}$ with the induced metric from $\mathbb{R}$. – Kolja Nov 22 '18 at 12:03
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    The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed. – Jyrki Lahtonen Nov 22 '18 at 12:04
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    Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two. – Kolja Nov 22 '18 at 12:06
  • That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen – nicomezi Nov 22 '18 at 12:09
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    Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense. – Kavi Rama Murthy Nov 22 '18 at 12:16

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