7
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ [1,2] is not a valid set here. Any set must be within [0,1] $\endgroup$
    – mmarky
    Jun 20, 2016 at 17:03
  • 1
    $\begingroup$ 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

17
$\begingroup$

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

$\endgroup$
14
$\begingroup$

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

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

$\endgroup$
1
  • 6
    $\begingroup$ $\{\frac 1 n\}$ is the archetype of a convergent sequence, so everyone thinks of it. $\endgroup$ Jun 20, 2016 at 23:00
8
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.