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

  • $\begingroup$ Maybe you can let $A$ be the rationals and see what happens. $\endgroup$ – Fabio Somenzi Aug 3 '19 at 20:08
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ @David I have found a better choice. $\endgroup$ – Display name Aug 3 '19 at 23:44
  • $\begingroup$ 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 $B=\{1/n: n\in \Bbb N\}.$ $\endgroup$ – DanielWainfleet Aug 4 '19 at 0:29
  • $\begingroup$ @DanielWainfleet I have found a new, albeit rather roundabout argument. $\endgroup$ – Display name Aug 4 '19 at 7:57
  • $\begingroup$ 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. $ $\endgroup$ – DanielWainfleet Aug 4 '19 at 11:42
  • $\begingroup$ @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. $\endgroup$ – Display name Aug 4 '19 at 23:29

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