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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.

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    $\begingroup$ It is the empty set! Is the empty set open? $\endgroup$
    – Thompson
    Aug 12, 2016 at 1:16
  • $\begingroup$ @ThompsonThe open ball is the set of all element x in Reals st $d\left ( x,x \right )<\epsilon $. Indeed, all elements in $\mathbb{R}$ satisfies this. But we require the open ball to be a proper subset of $\mathbb{R}$. $\endgroup$ Aug 12, 2016 at 1:22
  • $\begingroup$ It's not clear to someone trying to help you what definitions you are allowed to use. For example, do you have a definition of a closed set that is not "complement is open"? $\endgroup$
    – Thompson
    Aug 12, 2016 at 1:25
  • $\begingroup$ @Thompson I'm using only the definition of open set, closed set and complement. $\endgroup$ Aug 12, 2016 at 1:26
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    $\begingroup$ Then the answers given are the right route for you: The condition to be open is vacuously true for the empty set. $\endgroup$
    – Thompson
    Aug 12, 2016 at 1:36

2 Answers 2

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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.

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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.

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