Proving the every subset of $M$ is clopen.

Question

Let $M$ be a metric space with the discrete metric, or more generally a homeomorph of $M$.

How can I prove that every subset of $M$ is clopen?

Answer 1

In a metric space, a set $A\subseteq M$ is open if for every $x\in A$, there exists some $r>0$ such that for any $y\in M$, $\rho(x,y)<r$ implies $y\in A$.

Now consider any set $A\subseteq M$. If $A$ is empty, we're done (the empty set is open by fiat). If $A$ is nonempty, consider any $x\in A$. Let $r=1/2$. Now, if $y\in M$ and $\rho(x,y)<1/2$, then $y=x$, since $\rho(x,y)=1$ for any $y\in M$ such that $y\neq x$. Consequently, $y\in A$ and $A$ is open.

(Intuition: In the discrete metric, the only neighborhoods of any point are (1) the singleton containing only that point; and (2) the whole space. Therefore, any point in any set trivially contains a neighborhood of the point, the singleton containing only that point. This property makes any set open.)

Therefore, any subset of a discrete metric space is open. Specifically, the complement of every set is open as well, which implies (by definition) that every set is closed, too. $\blacksquare$

Answer 2

Hint: Let $A\subset M$, then $A$ is open. Now it's obvious that $M$ \ $A\subset M$, thus $M$ \ $A$ is open$\implies A$ is closed.

Answer 3

Define the discrete metric as a metric $d(·, ·)$, such that for any $x, y ∈ M$, $d(x, y) = 0$ if $y = x$ or 1 if $y \neq x$. Let $S ⊂ M$ be an arbitrary subset of the metric space.

If $S = ∅$ or $S = M$,we have by definition that $S$ is clopen. Otherwise, S is open if for any $x ∈ S$, there exists $r > 0$ such that $B(x, r) ⊂ S$. Fix an $x ∈ S$ and let $r = 1/2$. Then $B(x,1/2)={y∈M : d(y,x)<1/2}={y∈M : y=x}={x}⊂S$.

Therefore any subset $S ⊂ M$ is open, which implies that the subset $T = M\S ⊂ M$ is also an open subset. Since the complement of $S$ is open, $S$ is by definition closed. Furthermore because $S$ is both open and closed, it is clopen.