$B\in\Bbb R$ has the property that given $b∈B$ there exists $k>0$ such that if $0<|b−x|A subset $B$ of $\mathbb R$ has the property that given $b ∈ B$ there exists $k > 0$ such that if $0 < |b − x| < k$ for some $x ∈ \mathbb R$, then $x \notin B$. Is $B$ countable?
I tried using the diagonal argument here to show that $B$ is uncountable but I can't seem to make much progress...
 A: The property implies that for every $b \in B$ there exists $k_b > 0$ such that $\langle b - k_b, b + k_b\rangle \cap B = \{b\}$.
Pick a rational number $q_b \in \left\langle b - \frac{k_b}2, b + \frac{k_b}2\right\rangle$ and consider the map $f : B \to \mathbb{Q}$ given by $f(b) = q_b$.
Take $b,c \in B$ such that $q_b = q_c$. Assume $k_c \le k_b$. We have $$|b - c| \le |b - q_b| + |q_b - q_c| + |q_c - c| < \frac{k_b}2 + \frac{k_c}2 \le k_b$$
so $c \in \langle b - k_b, b + k_b\rangle \cap B = \{b\}$ which implies $c = b$.
Therefore, $f$ is injective. Since $\mathbb{Q}$ is countable, we conclude that $B$ is at most countable.
A: The hypothesis clearly implies that $B$ has no limit points. By Bolzano - Weirstarss Theorem $B$ can have only finitely many points in $(-n,n)$ for any $n$. so $B$ is at most countable.
A: Im not sure I understand your question.
Take interval $(0,2)= B\subseteq \Bbb R$ and let $b=1\in B$. Take $k=4>0$.  Naturally, for an $x=4\in\Bbb R$, we have $0<|1-4|=3<4$ .  Clearly $4\not\in B$ and $B$ is uncountable.
