Let $C \subseteq [0,1]$ be uncountable, show there exists $a \in (0,1)$ such that $C \cap [a,1] $ is uncountable

From what I know so far if something is countable then it has the same cardinality as $\mathbb{N}$, so to show there exists an $a \in (0,1)$ that works then I need to show that $\mid\,\mathbb{N}\mid \, < \,\mid C \cap [a,1] \,\mid$

If I assume $C$ is uncountable but $\neg\exists a \in (0,1)$ s.t. $C \cap [a,1]$ is uncountable. Then there is a surjection from $\mathbb{N}$ to $C \cap [a,1]$ and an injection from $C \cap [a,1]$ to $\mathbb{N}, \forall a \in (0,1)$

The union of countable sets are countable, so $\bigcup_{a=1}^{\infty} \bigg( C \cap [a,1] \bigg) $ is countable thus surjective from $\mathbb{N}$ and injective to $\mathbb{N}$.

The intuition makes perfect sense but I'm getting hung up on how I can get to my contradiction.

  • $\begingroup$ What do you mean by $\bigcup_{a=1}^{\infty}$? Isn't $a$ constrained to be between $0$ and $1$? $\endgroup$
    – user169852
    Sep 26, 2015 at 19:18
  • $\begingroup$ ah yes you are correct, I knew that was incorrect $\endgroup$ Sep 26, 2015 at 19:23

3 Answers 3


Suppose that $C\cap [\tfrac{1}{n}, 1]$ is countable for all $n$. Then

$$C\cap [0,1] = C\cap\big(\{0\}\cup \bigcup_{n=1}^\infty [\tfrac{1}{n},1]\big) = (C\cap \{0\}) \cup \bigcup_{n=1}^\infty (C\cap [\tfrac{1}{n}, 1])$$

would be countable too. Your intuition is perfectly correct.


Hint: $$ C \cap (0,1)=\bigcup_{n\in \Bbb N } C \cap [1/n,1] $$ Because the countable union of countable sets is countable, $C$ can only be uncountable if one of the sets in the union is uncountable.


Hint: Your ideas are good but you could try considering a countable union when a takes the values 1/n for n a natural number.


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