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let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is closed.

Can a finite subset of $X$ be open ?


Definitions:

  • a set $F\subseteq X$ is closed (in$(X,d)$) if $\bar F =F$.
  • a set $U\subseteq X$ is open (in$(X,d)$) if $U^o=U$
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4  
Hint: Can you show that any set containing just one element is closed? For the second question, consider the discrete metric. –  Tobias Kildetoft Mar 20 '13 at 17:40
    
For such basic questions it will help answerers if you include the precise definitions you have been given. What is your definition of an open set? A closed set? –  Pete L. Clark Mar 20 '13 at 17:42
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For the second question consider a finite metric space (just take $\{0,1,2\}$ with $\lvert \cdot-\cdot\rvert$ for example) –  Stefan Mar 20 '13 at 17:54
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Another way to read your definition of closed is that a set is closed in $(\mathbb{X},d)$ if and only if it contains all of its limit points. How many limit points does a finite set have? –  Todd Wilcox Mar 20 '13 at 18:01

3 Answers 3

up vote 5 down vote accepted

For an approach even more basic than Andrew Salmon’s, let $\langle X,d\rangle$ be a metric space, and let $F$ be any finite subset of $X$. The empty set is closed by definition, so we might as well assume that $F\ne\varnothing$. Now suppose that $x\in X\setminus F$, and let $r_x=\min\{d(x,y):y\in F\}$. Then $r_x>0$ (why?); what can you say about $B(x,r_x)$, the open ball of radius $r_x$ centred at $x$?

Yes, a finite set in a metric space can be open. First, the empty set is always open. Other than that, though, it depends on the space. No finite, non-empty subset of $\Bbb R^n$ is open, for instance, for any $n\in\Bbb Z^+$. However, if $X$ is any set at all, the function $d:X\times X\to\Bbb R$ defined by

$$d(x,y)=\begin{cases}1,&\text{if }x\ne y\\0,&\text{if }x=y\end{cases}$$

is a metric, often called the discrete metric, and every subset of $X$ is open.

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answering (why?) :$r_x>0$ because $F\cap X/F=\varnothing$.(Is this make any sense?) –  Jhwana Mar 20 '13 at 18:22
    
$B(x,r_x)$ is contained in $X/F$ thus $X/F$ is open , so $F$ is closed. –  Jhwana Mar 20 '13 at 18:27
    
@Fayz: I’m afraid not. Perhaps it will be easier if I let $F=\{x_1,\dots,x_n\}$ for some finite $n$. Then $r_x$ is the minimum of the numbers $d(x,x_k)$ for $k=1,\dots,n$. Since $x\notin F$, each $d(x,x_k)>0$. And there are only finitely many of these positive numbers, so there is a smallest one, which I call $r_x$. But your second comment is right on the money. –  Brian M. Scott Mar 20 '13 at 18:27
    
Thank you so much –  Jhwana Mar 20 '13 at 19:44
    
@Fayz: You’re very welcome. –  Brian M. Scott Mar 20 '13 at 19:46

For an answer that doesn't require any knowledge of separation axioms, consider the limit points of $\{ x \}$. Fix a point $y$ not in this set. Can $y$ be a limit point (hint: remember that $d(x,y) \ne 0$)?

Now remember that the finite union of closed sets is closed.

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Metric spaces are hausdorff spaces (T2) and therefore T1. This means that points are closed and a finite union of closed sets, as is well known, is closed.

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Good answer, +1 –  Rustyn Mar 20 '13 at 17:51

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