Claim: The empty set is open.
Proof. Assume that the empty set is closed. Then, there must be one point such that any point in its ball is not inside of the empty set. However, the empty set has no point at all and therefore no such point and ball exists. This is a contradiction.
Claim: $\mathbb{R}$ is open.
Proof. Assume $\mathbb{R}$ is closed. Then its complement, the empty set, is open. But, we know that the empty set is closed. This is a contradiction.
What I am worried about is that for the first proof, empty set is closed too.
So, getting contradiction while assuming that empty set is closed is little bit weird.
and same reason goes for second proof, since it is clear that R is closed too.
Can anyone modify it please?
Best Answer
'Not closed' does not mean open, for example the set $[0,1)$ is neither open nor closed. And, as you suggest, sets can be both open and closed (as both $\varnothing$ and $\mathbb{R}$ are).
You should use the definition of 'open' directly. Namely, for $X=\varnothing$ and $X=\mathbb{R}$, prove that for every $x \in X$ there exists $\varepsilon > 0$ such that $(x-\varepsilon, x+\varepsilon) \subseteq X$.