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Prove that if $S$ is a finite set then $S$ has no limit points.

Can someone tell me if my approach is correct:

Proof: Suppose $S$ is a finite set, then we can write $S = \{a_1, a_2, \ldots, a_n\}$ with $a_i \neq a_j$ if $i \neq j$. Suppose to the contrary that $S$ has a limit point $x_0$. Then by definition given any $\varepsilon > 0$ there exists $x \in S$ with $x \neq x_0$ such that $\vert x - x_0 \vert < \varepsilon$. Choose $\varepsilon$ to be the smallest distance between any two $a_i, a_j \in S$ with $i \neq j$. We can immediately see that there is no $x \in S$ such that $\vert x - x_0 \vert < \varepsilon$ holds, a contradiction. Thus we can conclude that $S$ has no limit points.

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Because it's good to think about this in multiple ways, here's another way to think about this problem. Prove: if $x$ is a limit point of the set $X$, then every neighborhood of $x$ contains infinitely many points of $X$. It's exactly the same thing (even has the same proof if you want go by contradiction), but it's a useful way to think about limit points. –  kigen Jan 17 '13 at 21:08
    
Thanks, I'm working through a book to review math analysis for this next semester and the next problem after this one is: Prove $x_0$ is a limit point of $S$ if and only if every neighborhood of $x_0$ contains infinitely many points of $S$. So I'll try that next! –  Robert Jan 17 '13 at 21:24
    
You have to be careful with your definitions to make this work. You need to have a reason why $\{0\}$ does not have the limit point $0$. if you nail that, then your proof should be straightforward. –  Mark Bennet Jan 17 '13 at 21:27
    
@MarkBennet I think the usual definition of limit point stipulates that every neighborhood contains a different point...though I have seen related definitions that do not require this. Henle's book uses the concept of a "near point" instead, where this requirement is lifted; granted, Henle's book is a rather poor resource for metric topology and abstract point-set topology, so I suppose the point is moot there. –  kigen Jan 17 '13 at 21:31
    
@proximal: There is some care to be taken about closed sets, boundaries and limit points to achieve consistent definitions - particularly in a historical context. I assume that the definition inherent in the question is as you suggest, but it is not given. And I think that the proof being asked for will be a consequence of the definition. –  Mark Bennet Jan 17 '13 at 21:49

3 Answers 3

up vote 7 down vote accepted

(You really should specify that $S\subseteq\Bbb R$; the statement isn’t true in topological spaces in general.)

Your argument is incomplete: it shows that no point of $S$ can be a limit point of $S$, but it doesn’t show that $S$ has no limit points in $\Bbb R$. Suppose, for instance, that $S=\{0,2\}$ and $x_0=1$. Then your $\epsilon=2$, and it’s not true that there is no $x\in S$ such that $|x-x_0|<2$; in fact, $|x-x_0|<2$ for both $x\in S$.

Your argument is fine if $x_0\in S$. If $x_0\notin S$, let $\epsilon=\min\{|x_0-a|:a\in S\}$. The distances $|x_0-a|$ are all positive, and there are only finitely many of them, so $\epsilon>0$, and it’s clear that there is no $x\in S$ such that $|x-x_0|<\epsilon$.

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When you wrote $\varepsilon = \min\{\vert x - a \vert : a \in S\}$ did you mean $x_0$ instead of $x$? And if I changed 'Choose $\varepsilon$ to be the smallest...' to 'Choose $\varepsilon$ to be the smallest distance between any $a \in S$ to $x_0$, that would fix the problem correct? –  Robert Jan 17 '13 at 21:21
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@Robert: Yes, I did; I’ve fixed it now. Your fix almost works: to cover the possibility that $x_0\in S$ you should let $\epsilon$ be the smallest non-zero distance from $x_0$ to any point of $S$. –  Brian M. Scott Jan 17 '13 at 21:28

Re-check your definition of a limit point. A limit point for $S$ does not necessarily have to be in $S$. That should lead you to the right answer.

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If $S=\{a_1,a_2\}\subseteq\mathbb{R}$, and $x$ lies halfway between $a_1$ and $a_2$, then your $\epsilon$ will not work.

Here is what works: Let $\epsilon$ be the smallest number in $\{|x-a_i|:i=1,\ldots,n,\,a_i\neq x\}$.

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Are we sure that $S\subseteq \Bbb R^n$? –  anon271828 Jan 17 '13 at 21:02
    
@anon271828 The OP used absolute-value notation. One can also interpret them as a norm if one wishes. –  Michael Greinecker Jan 17 '13 at 21:03
    
Yes I meant specifically $\mathbb R$ which is why I tagged real-analysis, but on hindsight I should have put it in the body of the problem as well. I'll be sure to specify next time! –  Robert Jan 17 '13 at 21:22
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@Robert You can be happy in knowing that eventually, exactly the same proof will work for fancy metric spaces. –  Michael Greinecker Jan 17 '13 at 21:23

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