# How to show those sets being countable or uncountable?

Consider the following questions, how do I show countability of set $$A$$ and $$B$$?

(a) A subset $$A$$ of $$\mathbb{R}$$ has the property that, given $$\varepsilon > 0$$ and $$x \in \mathbb{R}$$, there exist $$a,b \in \mathbb{R}$$ with $$a \in A$$ and $$b \not \in A$$, such that $$|x-a|<\varepsilon$$ and $$|x-b|<\varepsilon$$. Can $$A$$ be countable? Can $$A$$ be uncountable?

(b) A subset $$B$$ of $$\mathbb{R}$$ has the property that, for every $$b \in B$$, there exists $$\varepsilon > 0$$ such that for every $$x \in \mathbb{R}$$, $$0<|b-x|<\varepsilon$$ implies $$x \not \in B$$. Is $$B$$ countable?

Questions come from Cambridge Mathematical Tripos

• Maybe you can let $A$ be the rationals and see what happens. – Fabio Somenzi Aug 3 '19 at 20:08
• You need to visualize what those mean. 1) means that in any interval $(a,b)$ no matter how small there are $x,y \in (a,b)$ so that $x \in A$ and $y\not \in A$. There should be a VERY obvious countable set were that is true. 2) says for every $b$ in $B$ there an interval $(b-\epsilon, b+\epsilon)$ where $b$ is the only point of $B$ in the interval. Another way of putting it is that for every $b$ there is a finite distance to the nearest other element of $B$. Can such a set be uncountable? – fleablood Aug 3 '19 at 23:21

a. $$A=\mathbb{Q}, \mathbb{Q}^c = \mathbb{R}/\mathbb{Q}$$ work equally well, so $$A$$ can be either. Notice that the statement is invariant upon taking complements.

b. Suppose $$B$$ is uncountable. By the uncountable pigeonhole principle, some interval of the form $$[n, n+1], n \in \mathbb{Z}$$ contains uncountably many elements of $$B$$ (at most one interval may contain $$n$$ for any $$n \in \mathbb{Z},$$ so we do not need to worry about overlap). WLOG, suppose $$n=0$$ and let $$B' = B \cap [0,1].$$

For every $$b\in B' \subseteq B,$$ take the corresponding open interval centered at $$b,$$ and consider the collection $$C$$ of all such intervals. By definition, every element of $$C$$ contains one element of $$B';$$ denote this property $$(*)$$. By an analogue of the Bolzano-Weierstrass argument, we have a limit point $$b \in B',$$ which comes with a sequence $$b_1, b_2, \dots \to b.$$

Any open interval containing $$b$$ must contain some $$b_i,$$ contradicting $$(*)$$. Thus, $$B$$ is countable.

The argument for part b can probably be simplified.

• @David I have found a better choice. – Display name Aug 3 '19 at 23:44
• Your argument for Q (b) is flawed (although $B$ $is$ countable).... $\epsilon$ depends on $b$. There might not exist a single $\epsilon$ that works for all $b\in B.$ For example $B=\{1/n: n\in \Bbb N\}.$ – DanielWainfleet Aug 4 '19 at 0:29
• @DanielWainfleet I have found a new, albeit rather roundabout argument. – Display name Aug 4 '19 at 7:57
• Same flaw. You assume there is a pair-wise disjoint family $C$ of open intervals with each $c\in C$ containing just one $b\in B$ and $\cup C\supset B$ "by definition". Such $C$ exists only by some specific properties of $\Bbb R.$ For $x\in B$ let $r_x>0$ such that $B\cap (x-r_x,\,x+r_x)=\{x\}.$ Let $C=\{(x-r_x/2,\,x+r_x/2):x\in B\}.$ Now $C$ is pair-wise disjoint. $\Bbb Q$ is dense in $\Bbb R$ so for $c\in C$ let $f(c)\in c\cap \Bbb Q.$ If $B$ were uncountable then $f$ would be injective from the uncountable set $C$ into the countable set $\Bbb Q.$ – DanielWainfleet Aug 4 '19 at 11:42
• @DanielWainfleet You are right. Disjointness is not guaranteed. Rather, what is guaranteed is that the intervals do not contain multiple elements of $B.$ Luckily, this weaker property is enough to derive a contradiction. – Display name Aug 4 '19 at 23:29