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Question: Show that $\mathbb{R}$ is a closed subset of $\mathbb{R}$.
$\mathbb{R}\setminus \mathbb{R}=\left \{ x \in \mathbb{R} \mid x\notin\mathbb{R} \right \}.$
I need to show that $\forall x \in \mathbb{R}\setminus \mathbb{R}, \exists \epsilon >0$ s.t $B_{\epsilon }\left ( x \right )\subseteq \mathbb{R}\setminus \mathbb{R}$.
But, the complement of $\mathbb{R}$ itself doesn't make sense. Or rather, for it to make sense, it must be an empty set.
I think you may be unfamiliar with the concept of "vacuous truth." If you say "Every element of the empty set has property X," that statement is automatically true, since there is no element of the empty set! This is called a vacuously true statement. $\mathbb{R}$\ $\mathbb{R}= \emptyset$. Can you prove the theorem now?
As a side note, you shouldn't use this to prove the theorem but it is required of any topological space $Y$ that $Y$ itself be closed.
A sentence beginning by $\forall x\in\emptyset$ is always true (because you can't find any counterexample!). Thus $\Bbb R\setminus\Bbb R=\emptyset$ is open in $\Bbb R$ and so $\Bbb R$ is closed in itself.
