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Prove: the empty set $\emptyset$ and $\mathbb{R}$ are closed.

Definition: Let $S \subset \mathbb{R}$.

  1. S is said to be open if every point of S is an interior point of S.

  2. S is said to be closed if and only if $\mathbb{R} \setminus S$ is open.

Proof: $\mathbb{R}$ is closed since $\mathbb{R} \setminus \mathbb{R}$ = $\emptyset$ is open

$\emptyset$ is closed since $\mathbb{R}$ \ $\emptyset$=$\mathbb{R}$ is open.

Is it right?

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    $\begingroup$ While $\emptyset$ and $\mathbb{R}$ are closed, they are also open. $\endgroup$ Mar 19, 2020 at 13:30

1 Answer 1

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$\mathbb{R}$ is closed since $\mathbb{R} \setminus \mathbb{R}$ = $\emptyset$ is open

While it is true that $\emptyset$ is trivially clopen (both open and closed), it would help to understand why by applying the definition

S is said to be open if every point of S is an interior point of S.

This means that: $\forall \ x \ \in S, \ \exists r \in \mathbb{R^+}, \ b(x;r) \subset S$, where $b(x;r)$ is the open ball of radius $r$ centered at $x$.

Since $\emptyset$ has no $x$, the above statement is trivially true and thus $\emptyset$ is open.

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