Why is the empty subset of a metric space always connected ?

How can i show that the empty subset of a metric space,X always connected ?

It is empty and so will have and so will have an empty boundary.That doesn't seem to be enough.

Also, my book says that, no other finite set can be connected,I don't really understand this, because every finite set will be a non empty proper subset of X and will always be closed (and never both open and closed) so shouldn't it be connected ?

Any help will be appreciated.

Re: Why is the empty subset of a metric space always connected ?

The empty subset trivially satisfies the conditions of connectedness. Can you represent the empty subset as the disjoint union of two nonempty open sets?

Let $\displaystyle x \in X$. Then $\displaystyle \{x\}$ is also connected (I don't know why your book would not include singleton sets. They are also connected). Again, it cannot be represented by the disjoint union of two or more nonempty open sets.

Let $\displaystyle A \subseteq X$ with $\displaystyle 2 \le \text{card}(A) \in \mathbb{N}$. Let $\displaystyle D = \{d(a,b) \mid a,b \in A, a\neq b\}$. $\displaystyle D$ must be a finite set since $\displaystyle A$ is. So, let $\displaystyle d = \min(D)$. Then for any $\displaystyle a \in A$, $\displaystyle B\left(a;\dfrac{d}{2}\right) = \{a\}$ is an open subset of $\displaystyle A$. Since every set containing a single point is an open subset of $\displaystyle A$, if $\displaystyle A$ has more than one point, let $\displaystyle U = \{a\}$ and $\displaystyle V = A \setminus \{a\}$. By induction, you can show that $\displaystyle V$ is open and nonempty. And $\displaystyle A = U \cup V$.