If $E$ is a subset of $\mathbb{R}$ that does not contain any of its limit points, then $E$ is countable.

Question

The set $E$ cannot be open. If open, then for some $(a,b)\subset E \implies c\in(a,b) $ is a limit point of $E$, a contradiction. Hence, the set must be closed. If $E=\emptyset$, then the conclusion is evident.

The set $E$ if bounded, cannot be infinite. Since, by Bolzano-Weierstrass theorem, $E$ must have a limit point, and being closed, that point must be in $E$. The conclusion in this case is also established.

Now, let $E$ be unbounded. Assume it is uncountable. We can pick a bounded uncountable subset $C$ of $E$. Hence by Bolzano-Weierstrass theorem, it must contain the limit point of some subsequence of an infinite bounded sequence in $C$. $E$ being closed, the limit point must belong to $E$. Therefore, we get a contradiction.

The theorem is established.

Is the proof correct? I looked up this question here but I did not get a satisfactory (they were a bit harder to grasp) answer. Please do check.

Answer 1

$E$ is a subset of a separable metric space. So, as a metric space in its own right, $(E,d)$ is separable. By hypothesis, $(E,d)$ has no accumulation points, therefore $E$ is the only dense subset of $(E,d)$: otherwise $x\in E\setminus(\text{a dense subset})$ would be an accumulation point of the dense subset. Since $(E,d)$ is separable, $E$ must be countable.

Answer 2

Suppose $E$ is uncountable. For each $x\in E$, there is an open interval $I_x$ centered at $x$ such that $I_x\cap E=\{x\}.$

Let $E_n=\{x\in E:|I_x|>\frac{1}{n}\}$. If each of these sets were countable then their union, $E$, would also be countable. So, there is an integer $N$ such that $E_N$ is uncountable.

Now for each $x\in E_N$, consider the interval $J_x$ centered at $x$, of length less than $\frac{1}{2N}$ and suppose that $J_x\cap J_y$ is nonempty for some $x\neq y$. Then, $x\in I_y$ (and $y\in I_x$), which is a contradiction.

Thus, the intervals $\{J_x\}_{x\in E_N}$ are a pairwise disjoint $uncountable$ collection, which contradicts the density of $\mathbb Q$ in $\mathbb R$.

Answer 3

In this blog post it is shown that $\Bbb R$ is hereditarily Lindelöf which is equivalent to:

every uncountable subset of $\Bbb R$ contains one of its limit points.

By the contrapositive your statement immediately follows.

Answer 4

For each $x\in E$ let $r(x)>0$ such that $E\cap (-r(x)+x,\,r(x)+x)=\{x\}.$

$r(x)$ exists for each $x\in E.$ Otherwise there would exist some $x\in E$ such that for every $r>0$ there exists $y\in E$ with $0<|y-x|<r,$ which would make $x$ a limit point of $E$ and also a member of $E.$

For $x\in E$ let $I(x)=(-r(x)/3+x,\,r(x)/3+x).$

Observe that if $x,x'$ are unequal members of $E$ then $|x-x'|\ge r(x)$ and $|x'-x|\ge r(x')$ so $$0< r(x)/2+r(x')/2\le |x-x'|/2+|x'-x|/2=|x-x'|.$$

If $x,x'$ are unequal members of $E$ then $I(x), I(x')$ are disjoint. Otherwise , suppose $y\in I(x)\cap I(x').$ Then $|x-y|<r(x)/3$ and $|y-x'|<r(x')/3.$ But then $$0<r(x)/2+r(x')/2\le |x-x'|\le |x-y|+|y-x'|<r(x)/3+r(x')/3$$ implying $0<[r(x)+r(x')]/2<[r(x)+r(x')]/3,$ which is absurd.

For each $x\in E$ let $f(x)\in \Bbb Q\cap I(x).$ Then $f:E\to \Bbb Q$ is one-to-one because if $x,x'$ are unequal members of $E$ then $f(x)\in I(x)$ and $f(x')\in I(x')$ with $I(x)\cap I(x')=\emptyset,$ so $f(x)\ne f(x').$

If there exists a one-to-one function $f$ from a set $E$ into a countable set (such as $\Bbb Q$) then $E$ must be countable.

Answer 5

For any set $S$ let $|S|$ denote the (finite or infinite) cardinal of $S.$

For any topological space $X:$

(1). $w(X)$ (the weight of $X$) is the least infinite cardinal $k$ such that $X$ has a base (basis) $B$ with $|B|\le k.$

(2). $c(X)$ (the cellularity of $X$) is the least infinite cardinal $k'$ such that if $F$ is any pair-wise disjoint family of open subsets of $X$ (a.k.a. a discrete open family in $X$) then $|F|\le k'.$

(3). We have $w(X)\ge c(X).$ (See Appendix).

(4). If $E$ is a subspace of $X$ and $B$ is a base for $X$ with $|B|\le w(X)$ then $B'=\{b\cap E:b\in B\}$ is a base for $E$ with $|B'|\le |B|\le w(X),$ so $w(E)\le w(X).$

(5). $w(\Bbb R)=\aleph_0.$ If $E\subset\Bbb R$ and if none of the limit points of $E$ in the space $\Bbb R$ are members of $E$ then $ F=\{\{x\}:x\in E\}$ is a discrete open family in the subspace $E$ so $$|E|=|F|\le c(E)\le w(E)\le w(\Bbb R)=\aleph_0.$$

Appendix. Proof of (3). Let $F$ be an open cellular family in $X$ and let $F_0=F \setminus \{\emptyset\}.$ Let $B$ be a base for $X$ with $|B|\le w(X).$ For $f\in F_0$ let $\psi(f)\in B$ with $\emptyset \ne \psi(f)\subset f.$ Then $\psi:F_0\to B$ is $1$-to-$1$ so $|F_0|\le |B|\le w(X).$ And $w(X)$ is an infinite cardinal so $|F|\leq |F_0\cup \{\emptyset\}|\leq w(X).$

Remark. The density of $X$, denoted $d(X),$ is the least infinite cardinal $k''$ such that $X$ has a dense subset $D$ with $|D|\le k''.$ For any space $X$ we have $w(X)\ge d(X)\ge c(X)....$ But if $X$ is not metrizable then $X$ may have a subspace $Y$ with $d(Y)>d(X)$ and $c(Y)>c(X).$