Given any uncountable subset $S$ of the unit interval. Then $S$ clearly has an accumulation point and indeed uncountably many (which might also be a nice exercise). So my question is: Is there an accumulation point, that again lies in $S$?
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1$\begingroup$ This blog post may be of interest. $\endgroup$ – David Mitra Feb 21 '13 at 12:24
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$\begingroup$ Others have answered this, but I thought it worth mentioning that even more is true: All but countably many points of $S$ are accumulation points of $S.$ This is part of one version of the Cantor-Bendixson theorem. (A few seconds after writing this, I noticed that the "even more is true" result is in Brian M. Scott's answer.) $\endgroup$ – Dave L. Renfro Feb 21 '13 at 19:11
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Yes, every uncountable subset of $\mathbb{R}$ contains at least one of its accumulation points.
Suppose otherwise, that $S \subseteq \mathbb{R}$ is uncountable and contains none of its accumulation points. Then for each $x \in S$ there is an $\epsilon_x > 0$ such that $( x - \epsilon_x , x + \epsilon_x ) \cap S = \{ x \}$. Then there is an $n \in \mathbb{N}$ such that $S^\prime = \{ x \in S : \epsilon_x > \frac 1n \}$ is uncountable. Consider the family $$\{ ( x - \tfrac {1}{2n} , x + \tfrac{1}{2n} ) : x \in S^\prime \}.$$ It can be shown that this is an uncountable family of pairwise disjoint open subsets of $\mathbb{R}$ which contradicts that the countable set $\mathbb{Q}$ is a dense subset of $\mathbb{R}$.
answered Feb 21 '13 at 12:20
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Yes; you can even take it to be a complete accumulation point.
Let $\mathscr{B}$ be any countable base for the topology of $\Bbb R$, e.g., the set of open intervals with rational endpoints. Let $$C=\{x\in S:\exists B\in\mathscr{B}(x\in B\text{ and }B\cap S\text{ is countable})\}$$ be the set of points of $S$ having an open nbhd containing only countably many points of $S$. For each $x\in C$ let $B_x\in\mathscr{B}$ be such that $x\in B_x$ and $B\cap S$ is countable, and let $\mathscr{B}_0=\{B_x:x\in C\}$. Then $\mathscr{B}_0$ is countable, so $$C=\bigcup_{B\in\mathscr{B}_0}(B\cap S)$$ is countable as well, and every point of $S\setminus C$ is a complete accumulation point of $S$.
answered Feb 21 '13 at 12:21
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$\begingroup$ You seem to use $C$ to refer to two different sets, although the meaning is clear. $\endgroup$ – Mark S. Sep 28 '16 at 11:40
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$\begingroup$ @Mark: It’s the same set in both displayed expressions; the whole point is that the righthand sides of those two expressions are equal. $\endgroup$ – Brian M. Scott Sep 28 '16 at 16:02
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$\begingroup$ I see what you're saying. I'm not sure why I didn't see that before. $\endgroup$ – Mark S. Sep 28 '16 at 18:09
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If any accumulation point of $S$ does not lie in $S$, then $S$ is discrete; so for any $s \in S$, there is an interval $I_s$ of length $\ell_s>0$ such that $S \cap I_s = \{s\}$. Moreover, $\displaystyle \sum\limits_{s \in S} \ell_s \leq \ell g([0,1])=1$. So $S$ is at most countable. Indeed, since $\displaystyle \sum\limits_{s \in S} \ell_s$ is finite, then $\{s \in S : \ell_s > 1/n \}$ is finite, hence $\displaystyle S= \{s \in S : \ell_s >0 \}=\bigcup\limits_{n >0} \{s \in S : \ell_s > 1/n \}$ is at most countable.
