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I have a metric space $(X, d)$ and I am trying to prove that any finite subset $F = \{x_1,\ldots,x_n\} $ of $X$ is closed. What I have by now is a proof that a subset $F$ of a metric space $X$ is closed if and only if it contains all of its accumulation points. What I think is that if I prove that my set $F$ contains all of it's accumulation points, then $F$ would be closed, right? But I have problems in prooving that $F$ contains all of its accumulation points. If anyone could tell me if I am correct and help with the last proof, that would be great.

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  • $\begingroup$ Read the pdf after the Corollary 3.3.8, p. 87 . $\endgroup$ – Tony Piccolo Jan 13 '15 at 13:22
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Assuming that the $x_i$'s are different one from each other, take the minimum of $d(x_i,x_j)$ for $i\not=j$, noted $\varepsilon$. You obviously have $\varepsilon > 0$. Now take $\eta = \frac{\varepsilon}{2}$. Let $x$ be a limit of sequence $(u_k)_k$ of points of $F$. The sequence is convergent so it's Cauchy : there exists an $N$ such that when $k,l\geq N$, $d(u_k,u_l)\leq \eta$. Then if $k\geq N$ we have $d(u_k,u_N)\leq \eta$. But $u_N$ and the $u_k$'s are point of $F$, so by choice of $\eta$, $u_k = u_N$ for $k\geq N$, and therefore $x = u_N \in F$. $F$ is stable by convergence, so it's closed.

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  • $\begingroup$ Thank you! I have two questions about your answer if you could clarify: - why if $k\geq N$ we have $d(u_k,u_N)\leq \eta$ (should't it be $d(u_N,u_l)\leq \eta$)? And what does it mean that $F$ is stable? $\endgroup$ – Jim Jan 13 '15 at 14:41
  • $\begingroup$ 1) Being Cauchy, there exists an $N$ such that when $k,l\geq N$, $d(u_k,u_l)\leq \eta$. Fixing $l$ equal to $N$ and letting $k$ vary, you get that if $k\geq N$ we have $d(u_k,u_N)\leq \eta$. 2) By $F$ is stable by convergence, I mean that any sequence of elements in $F$ that is convergent has its limit in $F$. It is stable by the operation of taking limits of convergent sequences, to put is simply. And this stability is a criterion of closedness. Don't hesitate to validate the answer if you're ok with it. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 13 '15 at 14:46
  • $\begingroup$ Great. One more thing. How do we know that the set F has limit - you said 'Let x be a limit of sequence...'. I found here en.wikibooks.org/wiki/Real_Analysis/Metric_Spaces that 'a finite set has no limit points' and our set F is finite. $\endgroup$ – Jim Jan 13 '15 at 15:51
  • $\begingroup$ By "let $x$ be a limit..." I mean : let $(u_n)_n$ be sequence of elements of $F$ such that it converges to some $x$. And after, I want to show that $x$ is in fact in $F$. Showing this (that is, showing that all converging sequences of elements of $F$ have their limit in $F$) ensures the closedness of $F$. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 13 '15 at 15:54
  • $\begingroup$ But that means that a finite set F has limit points? $\endgroup$ – Jim Jan 13 '15 at 15:59
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Hint:Can you show that any set containing just one element is closed and the finite union of closed set is closed?

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  • $\begingroup$ I guess that is also a valid statement. Would this be a valid proof? Take $\{x\}$ which is subset of $X$. Now, consider any point $y \in X$ such that $y \neq x$. Then there exists (or for every?) $r > 0$ such that $d(x,y) < r$ where $r > 0$. We can take a ball around $y$ with radius $r$ such that $B(y, r)$ intersected with $\{x\}$ is an empty set. That means that the complement of $\{x\}$ is open so $\{x\}$ is closed? $\endgroup$ – Jim Jan 13 '15 at 20:07

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