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I'm stuck on the following real-analysis problem and could use a hint:

Consider $\mathbb{R}$ with the standard metric. Let $E \subset \mathbb{R}$ be a subset which has no limit points. Show that $E$ is at most countable.

I'm primarily confused about how to go about showing that this set $E$ is at most countable (i.e. finite or countable).

What I can show: since $E$ has no limit points, I can show that for every $x \in E$, there is a neighborhood $N_{r_x}(x)$ where $r_x > 0$ that does not contain any other point $y \in E$ where $y \neq x$. This suffices to show that every point within x is an isolated point.

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Hint: There must be at least one rational number in $N_{r_x}(x)$. – Thomas Andrews Sep 26 '12 at 17:42
up vote 4 down vote accepted

Hint. $\mathbb{Q}$ is dense in $\mathbb{R}$. Say there is an uncountable set without limit points. How is $\mathbb{Q}$ shared in their neighborhoods?

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For $n\geq1$ let $E_n:=E\cap[{-n},n]$. As $E=\bigcup_{n\geq1} E_n$, it is enough to show that each $E_n$ is countable, or even finite. Let an $n\geq1$ be given, and consider the set $I:=[{-n},n]$. Each $x\in I$ possesses an open neighborhood $U(x)$ that contains at most one point of $E$. The family $\bigl(U(x)\bigr)_{x\in I}$ is an open covering of $I$, and as $I$ is compact there are $N$ points $x_k\in I$ such that $I\subset\bigcup_{k=1}^N U(x_k)$. Since each $U(x_k)$ contains at most one point of $E$ it follows that $$\#E_n=\#(E\cap I)\leq\sum_{k=1}^N E\cap U(x_k)\leq N\ ,$$ as claimed.

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Each $x\in I$ possesses an open neighborhood $U(x)$ that contains at most one point of $E$... Why is that so? – Did Sep 26 '12 at 18:45
@did: Because $E$ has no limit points. And the OP has already shown this, according to the work in the question. – Brian M. Scott Sep 26 '12 at 19:28
@BrianM.Scott No. The fact that $E$ has no limit point does not imply this. And what the OP shows concerns only points in $E$, not every point in $I$. – Did Sep 26 '12 at 19:36
@did: You’re right on the second count but wrong on the first: if $E$ has no limit points, it must be both closed and discrete. – Brian M. Scott Sep 26 '12 at 19:39
@BrianM.Scott Are you saying that, if E has no limit points, then each x in I possesses an open neighborhood U(x) that contains at most one point of E? – Did Sep 26 '12 at 20:14

Note: The union of countably many countable sets is countable.

Assume $E$ is not countable. Then one of the sets $[n,n+1]\cap E$ with $n\in \mathbb Z$ is uncountable. Starting with $a_0=n$, $b_0=n+1$ we find a sequence of nested intervals $[a_k, b_k]$ such that $[a_k,b_k]\cap E$ is uncountable. In fact we can simply bisect an interval at each step and note that at least one of the halves must have iuncountably many points in common with $E$. In other words, we let $a_{k+1}=a_k$, $b_{k+1}=\frac{a_k+b_k}2$ if $[a_k, \frac{a_k+b_k}2]$ is uncountable and let $a_{k+1}=\frac{a_k+b_k}2$, $b_{k+1}=b_k$ otherwise (and observe that then $[a_{k+1},b_{k+1}]\cap E$ is uncountable as well. The nested intervals contain a point $c\in \mathbb R$. This $c$ has some (in fact uncountably many) points of $E$ in every $\epsilon$-neighbourhood. Indeed $[a_k,b_k]$ is contained in the $\epsilon$-neighbourhood as soon as $2^{-k}<\epsilon$.

Or: If $E\cap [0,\infty)$ and $E\cap(-\infty,0]$ are both countable, then so is $E$. Hence assume wlog that $E\cap [0,\infty)$ is uncountable. Let $a=\inf\{x\in \mathbb R\colon [0,x]\cap E\mathrm{\ is\ uncountable}\}$ If $a=\infty$ then $E\cap [0,\infty)=\bigcup_n E\cap[0,n)$ is the union of countable sets, hence countable. If on the other hand $a$ is finite, then $[a,a+\epsilon)\cap E$ is uncountable for any $\epsilon>0$, hence $a$ is a limit point.

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It’s enough to assume that $E$ has no limit points in $E$. Then for each $x\in E$ there is an $n(x)\in\Bbb N$ such that $(x-2^{-n(x)},x+2^{-n(x)})\cap E=\{x\}$; call this interval $U_x$. The rationals are dense in $\Bbb R$, so for each $x\in E$ we may choose a rational $q_x\in U_x$. Suppose that $E$ is uncountable. $\Bbb Q$ is countable, so there must be some $p\in\Bbb Q$ such that $E_0=\{x\in E:q_x=p\}$ is uncountable. And $\Bbb N$ is also countable, so there must be some $m\in\Bbb N$ such that $E_1=\{x\in E_0:n(x)=m\}$ is uncountable. But then $p\in U_x=(x-2^{-m},x+2^{-m})$ for each $x\in E_1$; equivalently, $|x-p|<2^{-m}$ for each $x\in E_1$.

Suppose that $x,y\in E_1$ with $p\le x<y$; then $x\in E\cap[p,y]\subseteq E\cap U_y$, contradicting the choice of $U_y$. A similar contradiction arises if there are $x,y\in E_1$ such that $x<y\le p$. Thus, $|E_1\cap[p,\to)|\le 1$ and $|E_1\cap(\leftarrow,p]|\le 1$, so $|E|\le 2$, contradicting the uncountability of $E_1$ and showing that $E$ must in fact be countable.

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Consider an uncountable subset $S$ of $\mathbb R^n $. Cover then $\mathbb R^n$ by the union:

$(\cup_{n=0,1,2,...}) B(0,n)$ , where $B(0,n)$ is the ball {$x$ in $\mathbb R^n$: $||x||<n$}. Then, by a standard cardinality argument (meaning: countable union of countables is countable), for some positive integer $n_0$ , the ball $B(0,n_0)$ will contain infinitely-many terms of $S$. Then use Weirstrass' theorem that says that a bounded, infinite subset of $\mathbb R^n $ has a limit point. So $S$ must contain a limit point.

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As you've already got, you can find these neighbourhoods that are just one point. Each one of these contains at least one rational, so there is at most countably many.

Reason why this doesn't prove any set is at most countable: the fact each neighbourhood only contains ONE point of $E$ (countably many is sufficient).

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proof by contradiction. suppose E is uncountable. consider the set E intersection with [n,n+1] for integers n. then there exist at least one n such that the above type of set is uncountable. clearly that set is bounded in real nos. then by Bolzano Weierstrass's theorem the above set has a limit point in R. a contradiction to the hypothesis.

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