# $\emptyset$, in a metric space, closed and open proof

Let $(X,d)$ be a metric space.

a)$\emptyset$ is open in $X$

b)$\emptyset$ is closed in $X$

a) By the definition of open set we have $A\in X$ is open if $B(x,\epsilon)\subset A$ for $x\in X$.The $\emptyset$ is open because a ball centred in the $\emptyset$ is contained in the set itself.

Questions:

How do I prove b)? Is my proof of $A$ right?

Is the $\emptyset$ opened and closed? I am confused on this issue.

• First of all: I think your b) should read, that the empty set is closed in $X$? And i don't understand your proof of a). You don't really say, why you can always find an $\epsilon$ for each $x$, such that the $\epsilon$-ball is in the set. Also: What is you definition of a closed set? Yes, the empty set is closed and open. – Verdruss Sep 27 '17 at 11:52
• @Verdruss I cannot find a ball in $\emptyset$ because it is empty but I guess I can speak theoretically of a ball centred in the $\emptyset$. Please check my update! Thanks for the feedback! – Pedro Gomes Sep 27 '17 at 12:04

Your proof of a) is not correct ! Suppose that $\emptyset$ is not open. Then there must be some $x \in \emptyset$ such that $B(x,\epsilon)$ is not contained in $\emptyset$ for each $\epsilon >0$. But $x \in \emptyset$ is impossible. Contradiction.

b) $\emptyset$ is closed, since the complement of $\emptyset$ is $X$, and $X$ is open.

Your proof of (a) seem to be confused, possibly due to inadequate understanding of the definition of open:

A set $E$ is said to be open if for any $x\in E$ there's an $\epsilon>0$ such that $B(x,\epsilon)\subseteq E$.

Since there are no elements in $\emptyset$ it's trivially true that for any element of $\emptyset$ whatever is true, especially that $B(x,\epsilon)\subset \emptyset$.

For (b) you also need the definition of closed, which is that the complement with respect to the space is open. The complement of $\emptyset$ is the entire space. And you only need to prove that for every point there is a $B(x,\epsilon)$ contained in the space, but this is trivially true by the definition of an open ball.