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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.


Context: Principles of Mathematical Analysis, Rudin

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    $\begingroup$ Know we use the fact that $\emptyset \subset E$ (the set of limit points is the empty set). $\endgroup$ Commented Jun 8, 2015 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. $\endgroup$ Commented Jun 8, 2015 at 21:53
  • $\begingroup$ I mean a metric space. $\endgroup$
    – RFZ
    Commented Jun 8, 2015 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. $\endgroup$ Commented Jun 8, 2015 at 21:56
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    $\begingroup$ I am studying analysis fron Rudin's book. I didn't know that every singletone is a closed set. $\endgroup$
    – RFZ
    Commented Jun 8, 2015 at 22:01

7 Answers 7

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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 $\endgroup$
    – RFZ
    Commented Jun 8, 2015 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 !! $\endgroup$ Commented Jun 8, 2015 at 22:04
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    $\begingroup$ You know in Rudin's book the convergent series is after that closed sets. $\endgroup$
    – RFZ
    Commented Jun 8, 2015 at 22:09
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    $\begingroup$ @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 $B_{r/2}(y)=\{z \in M: d(z,y)<r/2\}$ $\endgroup$ Commented Jun 8, 2015 at 22:21
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    $\begingroup$ @AlonsoDelfín Actually you could use $r$ instead of $r/2$ in that proof $\endgroup$ Commented Jun 8, 2015 at 23:15
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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). Another way to say this is that the empty set is a subset of any set.

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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. $\endgroup$ Commented Jun 8, 2015 at 22:12
  • $\begingroup$ Why you take $a\notin A$? $\endgroup$
    – RFZ
    Commented Jun 8, 2015 at 22:18
  • $\begingroup$ $a \in \overline X \iff d(a, X) = 0$. $\endgroup$ Commented Jun 8, 2015 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? $\endgroup$ Commented Jun 9, 2015 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
    Commented Jun 9, 2015 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
    Commented Jun 9, 2015 at 15:29
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Note that $A = \bigcup_{i=1}^n \{x_i\}$. Also each $\{x_i\}$ is closed since $\Bbb R \setminus \{x_i\} = (-\infty,x_i)\,\cup (x_i, \infty)$ and both $(-\infty,x_i)$, $(x_i, \infty)$ are open in usual topology on $\Bbb R$ (You can also check this without invoking topology by considering the fact that any point in any of these subsets has a open neighbourhood lying in that subset and hence, every point of these subsets is an interior point. So, both of these are open).

Also, we know that the finite union of closed sets is closed and hence every finite set is closed in a metric space.

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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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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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