3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Prove that every singleton set is open. $\endgroup$ Sep 17 '13 at 9:44
2
$\begingroup$

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. $\blacksquare$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ What is $X$???? $\endgroup$
    – Don Larynx
    Sep 17 '13 at 9:59
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.