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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.

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  • $\begingroup$ 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. $\endgroup$ – Verdruss Sep 27 '17 at 11:52
  • $\begingroup$ @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! $\endgroup$ – Pedro Gomes Sep 27 '17 at 12:04
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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.

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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.

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To add to the other answers:

If your definition of a closed set is that every point of the boundary is contained in the set itself or that every limit point of a sequence in the set is contained in the set, you need to adjust the given proofs a little bit. But it will be always due to the fact, that there are no points in the empty set (hence the boundary is empty and there are no sequences in the set).

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