1

Let be $E,F\subset \Bbb{R}$ two subsets such that $E$ is uncountable and $F^c$ is countable. Is $E\cap F$ uncountable?

I guess it is true, but I am not sure since I don't see a way in order to prove it or any counterexamples.

I have tried to prove it realizing that $F$ is dense at the usual topology, but I don't figure 0ut how continue on. Any possible help would be appreciated.

Superdivinidad
  • 635
  • 10
  • Basically, if we take the difference between the cardinality of $\mathbb{R}$ and the cardinality of $\mathbb{Q}$, we get a set whose cardinality is greater than the cardinality of $\mathbb{Q}$ itself (which is equal to $\aleph_0$). Hoping that this can help you... – Marco Ripà Sep 26 '23 at 10:27
  • 2
    I would do it slightly differently to Ramiro. $F\cap E = \left(\mathbb{R}\setminus F^c \right) \cap E = \left( \mathbb{R} \cap E \right) \setminus \left( F^c \cap E\right)$ = uncountable - countable = uncountable. – Adam Rubinson Sep 26 '23 at 10:43

1 Answers1

4

Yes, if $E$ is uncountable and $F^c$ is countable, then $E \cap F$ is uncountable.

Proof: Note that $E = (E \cap F) \cup (E \cap F^c)$. Since $F^c$ is countable, we have that $E \cap F^c$ is countable. Since $E$ is uncountable, we must have that $E \cap F$ is also uncountable (otherwise $E$ would be countable).

Ramiro
  • 17,521