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

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.

Solution 1:

(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 $|x-x_0|<\epsilon$.

Solution 2:

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.