If every accumulation point of a set $X\subset \mathbb R$ is unilateral, then $X$ is countable.

Question

When referring to unilateral accumulation points we're talking about either

$X'_+ = \{a\in\mathbb R : \forall \epsilon>0 $ $(a,a+\epsilon) \cap X \neq \emptyset \}$

or

$X'_- = \{a\in\mathbb R : \forall \epsilon>0 $ $(a-\epsilon,a) \cap X \neq \emptyset \}$

Answer 1

For every accumulation point $a$ of $X$, we know that there exists $\epsilon>0$ such that $(a,a+\epsilon)\cap X=\emptyset$ (because $a$ is left, not right unilateral) or $(a-\epsilon,a)\cap X=\emptyset$ (because $a$ is right, not left unilateral). In the first case, associate $a$ with a rational number $q(a)$ in $(a,a+\epsilon/2)$, in the second case in $(a-\epsilon/2,a)$. Then $q(a)\ne q(a')$ if $a,a'$ are distinct accumulation points. We conclude that there are only countably many accumulation points. Similarly, we see that $X$ has only countably many isolated ponts. As all non-isolated points of $X$ are also accumulation points, the claim follows.

Answer 2

Let $X$ be an uncountable subset of $\Bbb R.$ We can show a stronger result: There exists an uncountable $Y \subset X$ such that for all $r>0$ and all $y\in Y$ the sets $(y,y+r)\cap Y$ and $(-r+y,y)\cap Y$ are uncountable, and such that $X\setminus Y$ is countable.

(i). Let $A$ be the set of open real intervals with rational endpoints. Let $B=\{b\in A: |b\cap X|\leq \aleph_0\}.$ Let $C$ be the unique countable family of pair-wise disjoint convex open real subsets such that $\cup B=\cup C.$

Let $Y=X\setminus (\cup \{\overline c: c\in C\}).$

(ii). Since $\cup \{\overline c: c\in C\} =(\cup C)\bigcup (\cup \{\overline c \setminus c:c\in C\})=(\cup B)\bigcup (\cup \{\overline c\setminus c:c\in C\}), $ we have $$X\setminus Y =(\cup \{b\cap X:b\in B\})\bigcup (\cup \{(X\cap (\overline c\setminus c): c\in C\}).$$ Now $B$ is countable and $b\cap X$ is countable for each $b\in B.$ Also $C$ is countable and $\overline c\setminus c$ contains at most $2$ points for each $c\in C.$ Therefore $X\setminus Y$ is countable. And so $Y$ is uncountable.

(iii). For any $V\subset \Bbb R,$ if $X\cap V$ is uncountable then $Y \cap V$ is uncountable, because $Y\subset X$ and $X\setminus Y$ is only countable.

(iv). If $y\in Y$ and $r>0$ such that $|(y,y+r)\cap Y|\leq \aleph_0$ then by (iii), $|(y,y+r)\cap X|\leq \aleph_0,$ implying $(y,y+r)\subset \cup B.$ (See footnote.) Then for some $c\in C$ we have $(y,y+r)\subset c,$ (See footnote again), implying $y\in \overline c$, which is absurd by the definition of $Y.$

Similarly if $y\in Y$ and $r>0$ and $|(-r+y,y)\cap Y|\leq \aleph_0,$ we also have an absurdity.

Thus $Y$ is an uncountable, co-countable subset of $X$ with the property that for any $y\in Y$ and any $r>0$ the sets $(y,y+r)\cap Y$ and $(-r+y,y) \cap Y$ are uncountable. (A fortiori, for any $y\in Y$ and any $r>0$ the sets $(-r+y,y)\cap X$ and $(y,y+r)\cap X$ are uncountable.)

Footnote: If $r>0$ and $|(y,y+r)\cap X|\leq \aleph_0$ then the set $S=\{(p,q): p,q\in \Bbb Q\land y\leq p<q\leq y+r\}$ is a subset of $B,$ so $(y,y+r)=\cup S\subset \cup B.$ By the definition of $C,$ if an open interval $I,$ such as $(y,y+r),$ is a subset of $\cup B,$ then $I\subset c$ for some $c\in C.$

Remark: Observe that $\cup B$ is the $\subset$-largest open set that has countable intersection with $X.$