I know that if $C \subseteq [0,1]$ is uncountable, then there exists $a \in (0,1)$ such that $C \cap [a,1]$ is uncountable. Is it still true for any infinite sets? That is, if $C \subseteq [0,1]$ is infinite, does there exist an $a \in (0,1)$ such that $C \cap [a,1]$ is infinite?

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    $\begingroup$ I'm not sure I understand your question, but $[0,1]\cap [1,2]=\{1\}$ is finite despite both being uncountably infinite... $\endgroup$
    – JMoravitz
    Jun 20, 2016 at 17:00
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    $\begingroup$ [1,2] is not a valid set here. Any set must be within [0,1] $\endgroup$
    – mmarky
    Jun 20, 2016 at 17:03
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    $\begingroup$ Out of interest, does the proof of this have to use (some version of) the Axiom of Choice, or is there a stronger proof than saying, "suppose that all the intersections are countable, then $C$ is a countable union of countable sets (namely, the union of all such intersections for some sequence of $a$ tending to $0$, plus possibly the set $\{0\}$) and hence is countable?". In that proof it's obvious why it breaks down replacing "uncountable" with "infinite", because a countable union of finite sets need not be finite. $\endgroup$ Jun 20, 2016 at 18:41
  • $\begingroup$ The title does not match the question. $\endgroup$
    – bof
    Jan 8, 2017 at 9:26

4 Answers 4


Not necessarily. Consider $C := \{\frac{1}{n} : n \in \mathbb{N}\}$.


Not true. Take $$C = \left\{1, \frac12, \frac13, \frac14, \ldots , \frac1N, \ldots \right\}.$$

[Jeez, everyone came up with the same example at once]

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    $\begingroup$ $\{\frac 1 n\}$ is the archetype of a convergent sequence, so everyone thinks of it. $\endgroup$ Jun 20, 2016 at 23:00

Every set E such that :

$$\forall \varepsilon >0, \quad E\cap(\varepsilon,1)\text{ is finite}$$

is a counter example.

You just need $0$ to be an accumulation point of your set.

You can take

$$\left\{\frac 1n ,\ n\in \mathbb N\right\}$$ for instance.


This argument is not true...a Counter example is the set:$$\left\{\frac 1n ,\ n\in \mathbb N\right\}$$


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