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Let $A\subseteq\mathbb{R}$ with $A$ uncountable.

Let $B\subseteq A$ where for every $b\in B$, $\exists n\in\mathbb{N}$ where $\left[b,b+\frac1n\right)\cap A$ is countable.

Now, show $B$ is countable.

I was thinking of assuming $B$ is uncountable and taking one interval for every $b\in B$. Then I would take a rational from each interval, and derive a contradiction as there are only countably many rationals. This doesn't work as the intervals could overlap. I'm unsure how to continue.

Andrés E. Caicedo
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James Gwin
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    Hint. Try using transfinite recursion up to $\omega_1$ and at each stage pick a rational number in the corresponding interval of some $b$ and remove all the other $b'$ that have this rational number in their interval. Note that the condition you have on $B$, assures that at each stage, you are throwing away at most countably many elements. – Shervin Sorouri Apr 26 '20 at 21:15
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    Also if you want to approach this problem without using recursion, you can go with this route: First fix some $n_b$ for each $b\in B$ which satisfies the condition on $b$ and let $I_b = [b, b +\frac{1}{n_b})$. Now note that each rational $q$ lies in at most countably many $I_b$'s so let $f_q:\mathbb{N}\rightarrow B$, enumerate those $b$'s for which $q\in I_b$. Now build a surjection from $\mathbb{Q}\times \mathbb{N}$ onto $B$, by letting $g(q, n) = f_q(n)$. I hope you can fill in the details. – Shervin Sorouri Apr 26 '20 at 21:28
  • If you have some topological background, you can first show that if $B$ were countable, this answer shows that there would be a point $b\in B$ such that $[b,b+\epsilon)\cap B\subseteq[b,b+\epsilon)\cap A$ is uncountable for every $\epsilon>0$, contradicting the hypothesis on $B$. (And you don’t even really need any topology to follow bof’s proof at the link embedded in that answer.) – Brian M. Scott Apr 26 '20 at 21:50

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