The question and its answer is given in the following picture:
Question 52.
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.Solution. Option (B) is true. Here's why: Recall that $\mathbb Q$ is dense in $\mathbb R$, i.e. $\operatorname{cl}(\mathbb Q) = \mathbb R$ where $\operatorname{cl}$ denotes the closure of a set. It follows that if $A$ is closed $\operatorname{cl}(\mathbb Q) = \mathbb R\subseteq \operatorname{cl}(A) = A \subseteq \operatorname{cl}(\mathbb R) = \mathbb R$. Thus $A = \mathbb R$.
Let's see why the others are lies. To see why (A) isn't true, consider $A = \mathbb R\setminus\{\pi\}$; the set $A$ is open because for each $x$ in $A$, the open ball centered at $x$ with radius $|x-\pi|/2$ is contained in $A$. To disprove (C), consider $A:= \mathbb Q\cup (0,1)$. The same counterexample disproves (D). Choice (E) is disproved by $A =\mathbb Q$, because $\sqrt 2$ is a limit point.
I am convinced with the solution and its justification. But the deletion of the wrong choices is not so clear for me as the interpretation is said in brief, could anyone explain this in detail for me please?
To disprove all other options, a counterexample is given.
The set $\mathbb R \setminus \{\pi\}$ is an open set containing each rational nummber as its complement ($\{\pi\}$) is closed. However, it is not equal to $\mathbb R$. The disproves $(A)$.
The set $\mathbb Q \cup (0,1)$ is an uncountable set containing each rational number. However, it is not equal to $\mathbb R$. The disproves $(C)$.
The set $\mathbb Q \cup (0,1)$ is an uncountable set containing each rational number. However, it is not open(as no point outside $(0,1)$ is an interior point.). The disproves $(D)$.
The set $\mathbb Q$ is a countable set containing each rational number. However, it is not closed as $\overline{\mathbb Q}=\mathbb R$. The disproves $(E)$.