If $A$ is a subset of the real line $\mathbb R$ and $A$ contains each rational number, which of the following must be true?
(a) If $A$ is open, then $A = \mathbb R$.
(b) If $A$ is closed, then $A = \mathbb R$.
(c) If $A$ is uncountable, then $A = \mathbb R$.
(d) If $A$ is uncountable, then $A$ is open.
(e) If $A$ is countable, then $A$ is closed.
My Approach:
(a) False. $A$ could be say $\mathbb R\setminus \{\sqrt{2}\}$ which is open as it is basically the union of two open sets $(-\infty, \sqrt 2) \cup (\sqrt 2, \infty)$.
(b) True. $\bar {\mathbb Q}$ is the smallest closed set containing $\mathbb Q$. And we know $\bar {\mathbb Q} = \mathbb R$. So if $A$ is closed, then $A = \bar A = \mathbb R$. It couldn't possibly be larger than $\mathbb R$.
(c) False. $\mathbb R \setminus \{\sqrt{2}\}$ contains $\mathbb Q$ but is uncountable.
(d) I'm not sure about this but I think it is false. Could someone give an explicit example of an uncountable set $A$ containing $\mathbb Q$ that is open?
(e) False. A counterexample is $\mathbb Q$ itself. We know that $\mathbb Q$ is neither open nor closed in $\mathbb R$.
