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A point $p$ in a metric space $X$ is said to be a condensation point of a set $E\subset X$ if every neighbourhood of $p$ contains uncountably many points of $E.$ [Note that $p$ need not be in $E.$]

Suppose $0$ is a condensation point of a set $E\subset\mathbb{R}.$ Let $\alpha>0.$ Then does $\{\alpha x: x\in E\}\cap E \neq \emptyset ?$

Something I know about condensation points is that the set of condensation points of an uncountable set is perfect.

Maybe the Cantor set with a well-chosen $\alpha$ is a counter-example?

Or maybe a counter-example can be constructed using the Axiom of choice?

Asaf Karagila
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Adam Rubinson
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Let $$E = \bigcup_{n = 1}^\infty (2^{-2n}, 2^{-(2n-1)}) = \left (\frac{1}{4},\frac{1}{2} \right ) \cup \left (\frac{1}{16},\frac{1}{8} \right ) \cup \dots$$ and $\alpha = 2$. Then $0$ is a condensation point for $E$, since any neighbourhood of $0$ contains one of these dyadic intervals, but $E$ and $2E$ are disjoint; $2E$ is (the interior of) $[0,1] \backslash E$.

Julius
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    What if $E$ is also dense in an interval around $0?$ – Adam Rubinson Sep 27 '24 at 14:38
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    @AdamRubinson With $\alpha = 2$ again, let $E$ be the set of all irrational $x \in (-1,1)$ with binary expansion $x=\pm 0.x_1x_2x_3\dots$, such that $x_{2n} = 0$ for all sufficiently large $n$. Then this is dense in $(-1,1)$, $0$ is a condensation point of $E$, and $2E$ is the set of all irrational $x \in (-2,2)$ whose odd digits in its binary expansion are eventually $0$, so $E$ and $2E$ are disjoint (if $x$ is such that the odd digits and even digits are both eventually $0$, then $x$ would be rational). – Julius Sep 27 '24 at 15:09
  • I agree that the set is dense in $(-1,1),$ and that $0$ is a condensation point of $E.$ But are you sure that $2E$ is the set of all irrational $x\in (-2,2)$ whose odd digits in its binary expansion are eventually $0?$ Take $0.001.$ Then $0.001\times 2 = 0.01$ or $0.0011111111\ldots,$ Either way, it equals to $\frac{1}{4},$ which is not irrational. Maybe you included the word irrational by mistake? – Adam Rubinson Sep 28 '24 at 15:19
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    @AdamRubinson I defined $E$ as a subset of irrational numbers, and so it wouldn't contain $0.001$. Really the intention is just to exclude trailing zeroes (or ones) which means we have to only consider irrationals. – Julius Sep 29 '24 at 19:48
  • My bad: I missed that you defined $E$ to be a set of irrational numbers. I think I get it now. Wow, that is so clever! – Adam Rubinson Sep 29 '24 at 21:51