2

In finite metric space all sets are open

Is it true because all singletons are open and the union of them is open too?

gbox
  • 12,867

2 Answers2

7

It's true, but another way:

In any metric space, finite or not, all singletons are closed. So finite sets are closed.

In a finite metric space, any subset has a finite (hence closed) complement. So all sets are open.

Henno Brandsma
  • 242,131
  • I read your proof here: https://math.stackexchange.com/questions/1548488/show-that-the-singleton-set-is-open-in-a-finite-metric-spce

    Basically the proof says that a singleton in finite metric space is open because for all point in the neighborhood of the singleton (and there are none) $N_{s}\subseteq {s}$?

     – gbox Aug 07 '17 at 21:42
  • @gbox The point of that proof was that the OP wanted to use the open-ball definition of openness in a metric space. So I defined a radius for an open ball. – Henno Brandsma Aug 07 '17 at 21:45
  • Yes, but in the end you showed that the ball only contains the singleton so why the ball is open? – gbox Aug 07 '17 at 21:47
  • 1
    @gbox Because $\forall p \in {x} \exists r > 0: B(p,r) \subseteq {x}$, which is the definition of open sets in a metric space. – Henno Brandsma Aug 07 '17 at 21:49
  • @gbox Or use the lemma: an open ball $(p,r)$ is an open subset of $(X,d)$ in any metric space. – Henno Brandsma Aug 07 '17 at 21:50
  • this is what I meant to say, thanks – gbox Aug 07 '17 at 21:50
1

Let $M=\{x_1,x_2,\cdots,x_n\}$ be a finite metric space.

Let $A=\{x_1\}$ Then $A^c=\{x_2,x_3,\cdots,x_n\}$ is a finite set, hence closed. Therefore, $A$ is open.

Similarly, all singletons are open.

Now, let $B$ be any subset of $M.$ Then $B$ is a finite union of open sets(singletons), hence open.

In general, all subsets of a metric space are open if singletons are open.

Edit 1: To show a finite set is closed, it is sufficient to show that any singleton is closed (since finite union of closed sets is closed)

Let $X$ be a metric space and $x \in X$. We need to show $\{x\}$ is closed. It suffices to show that $\{x\}^c$ is open. Let $ y \in \{x\}^c.$ Then $y\neq x.$ Hence, $d(x,y)>0.$ Let $$r=\frac{d(x,y)}{2}>0$$ Then $B(y,r) \subseteq \{x\}^c.$ This shows $\{x\}^c$ is open.

Edit 2: Let $z \in B(y,r)$. Then $$d(z,y)<r=\frac{d(x,y)}{2}<d(x,y)$$ Hence, $z \neq x$. Therefore, $z \in \{x\}^c.$ As $z \in B(y,r)$ was arbitrary, therefore $B(y,r) \subseteq \{x\}^c.$

Sahiba Arora
  • 10,847