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.
Best Answer
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.