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Question:

Prove that a finite subset in a metric space is closed.

My proof-sketch: Let $A$ be finite set. Then $A=\{x_1, x_2,\dots, x_n\}.$ We know that $A$ has no limits points. What's next?

Definition: Set $E$ is called closed set if $E$ contains all his limits points.

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    $\begingroup$ Know we use the fact that $\emptyset \subset E$ (the set of limit points is the empty set). $\endgroup$ – Noah Olander Jun 8 '15 at 21:52
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    $\begingroup$ What is the context? Subsets of $\mathbb R$ with the usual topology/metric? Without context it is impossible to answer your question as in fact in an arbitrary topological space not all finite sets need be closed. $\endgroup$ – Ittay Weiss Jun 8 '15 at 21:53
  • $\begingroup$ I mean a metric space. $\endgroup$ – ZFR Jun 8 '15 at 21:55
  • $\begingroup$ @IttayWeiss You are right. Although since the OP tag this as real analysis I assume he might be talking of any finite dimensional metric space, in which case the finite sets are always closed. $\endgroup$ – Alonso Delfín Jun 8 '15 at 21:56
  • $\begingroup$ @Pacman do you happen to know then that in a metric space every singleton is a closed set? and that the union of finitely many closed sets is closed? combine... $\endgroup$ – Ittay Weiss Jun 8 '15 at 21:57
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Thinking $A$ as a subset of a metric space $M$. An easy approach will be to use that the single points $\{x_j\}\subset M$ are closed (you know why?), then of course $$ A=\bigcup_{j=1}^n \{ x_j \} $$ Since $A$ is a finite union of closed sets, it is itself closed.

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  • $\begingroup$ I don't know that single point is closed $\endgroup$ – ZFR Jun 8 '15 at 22:03
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    $\begingroup$ Yes, in metric spaces they are. And it is very easy to prove PROOF: Take any convergent sequence in $\{ x \}$, then of course it converges to $x \in \{ x \}$, thus $\{x\}$ is closed !! $\endgroup$ – Alonso Delfín Jun 8 '15 at 22:04
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    $\begingroup$ You know in Rudin's book the convergent series is after that closed sets. $\endgroup$ – ZFR Jun 8 '15 at 22:09
  • $\begingroup$ @Pacman I see. Here is a proof without using the sequence criteria for closed sets. PROOF Take any $y \in M \setminus \{ x \}$ and fix $r=d(x,y)$, then clearly $B_{r/2}(y) \subset M \setminus \{ x \}$, thus $M \setminus \{ x \}$ is open and hence $\{ x \}$ must be closed $\blacksquare$ . Here $B_{r/2}(y)=\{z \in M: d(z,y)<r/2\}$ $\endgroup$ – Alonso Delfín Jun 8 '15 at 22:21
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    $\begingroup$ @AlonsoDelfín Actually you could use $r$ instead of $r/2$ in that proof $\endgroup$ – Mario Carneiro Jun 8 '15 at 23:15
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If $M$ is a metric space then every subset $A =\{x_1, \ldots, x_n\} \subseteq M$ is closed. In fact, if $a \notin A$ then $d(a,A)$ is the least of the numbers $d(a,x_1) ,\ldots, d(a,x_n)$ thus, $d(a,A) > 0$.

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  • $\begingroup$ In particular, $\{x\}$ is closed. $\endgroup$ – Aaron Maroja Jun 8 '15 at 22:12
  • $\begingroup$ Why you take $a\notin A$? $\endgroup$ – ZFR Jun 8 '15 at 22:18
  • $\begingroup$ $a \in \overline X \iff d(a, X) = 0$. $\endgroup$ – Aaron Maroja Jun 8 '15 at 22:20
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If we show that the compliment of $A$ is open, then $A$ is closed.

Let $B=A^c$, and suppose that $y \in B$. We need to show that there is an $r >0$ for which $B_r(y) \cap A = \emptyset$, that is $B_r(y) \subset B$.

If we let $r = \min \{ d(y,x_i) : i = 1,...,n\}$ then it must be true that $B_r(y) \cap A = \emptyset$, since for every $z \in B_r(y)$ we have $d(z,y) < d(y,x_i)$ for each $i=1,...,n$.

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  • $\begingroup$ Can you do this without a metric? $\endgroup$ – steven gregory Jun 9 '15 at 1:58
  • $\begingroup$ The metric defined your topology and therefore your open sets. In a general topology, you would not expect singletons or finite sets to be closed. $\endgroup$ – Joel Jun 9 '15 at 15:27
  • $\begingroup$ @StevenGregory, in any case, the explicit use of a metric can be avoided, but it will be implicit through the topology. $\endgroup$ – Joel Jun 9 '15 at 15:29
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I think the simplest answer to this, following Baby Rudin (Principles of mathematical analysis 3rd ed.) terminology, is:

The definition of a closed set is in 2.18.d: a set $A$ is closed if every limit point of $A$ is an element of it. However, none of the points $x_i \in A$ are limit points of $A$: for each $x_i$, pick $r = min(d(x_{i-1} x_i), d(x_i, x_{i+1}))$ i.e. the minimum distance between itself and its neighbours (to take care of corner cases, define $d(x_0, \_) = \infty$ , $d(\_, x_{n+1}) = \infty$). Now, any neighbourhood $N_{r'}(x_i)$ for $r' < r$ doesn't contain any other point of $A$ apart from $x_i$ so $x_i$ can't be a limit point. Therefore, $A$ is closed "vacuously" i.e. trivially, because the condition for it to be closed (all limit points of $A$ are elements of $A$) is trivially satisfied because the set of all limit points of $A$ is empty (a set of 0 points has any property).

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If some open set $G_i$ containing $x_1$ does not contain $x_i$, for $i=1,\ldots,n$, then $\bigcap_{i=1}^n G_i$ is an open set containing $x_1$ and not containing any of $x_2,\ldots,x_n$. Therefore $x_1$ is not a limit point of the set. The same argument applies to show that $x_2,\ldots,x_n$ are not limit points.

If the space is a Hausdorff space, then certainly some open set $G_i$ containing $x_1$ does not contain $x_i$ if $i\ne1$.

If it's a metric space, you can take $G_i$ to be the open ball of radius $d(x_1,x_i)$ centered at $x_1$.

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Let $(X, d)$ be a metric space and $A = \{ x_1, \ldots, x_n \} \subset X$ be finite. Then the following holds: A sequence $(a_n)_{n \in \Bbb N}$ in $A$ is convergent, if and only if there exists a $N \in \Bbb N$ and a $i \in \{ 1, \ldots, n\}$, such that $$ a_n = x_i \quad \text{for all } n \geq N \; ,$$ i.e. the sequence $(a_n)_{n \in \Bbb N}$ will be constant up to a index $N$. This means, that the limit points of $A$ are $x_1, \ldots, x_n$, so $A$ contains all its limit points, thus $A$ is closed.

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