Show that for a finite metric space A, every subset is open Let A be a finite metric space .I want to prove that every subset of A is open.
I let the set B, be any subset of A.
Since A is finite,then I know that A/B is also finite.I'm stuck here how can this help me reach to a proof? I beg your help
 A: Massive hint: In a metric space, finite point sets are closed. So suppose that you have a subset $B$ of $A$. Then $A \setminus B$ is a finite point set so.....
A: A space is discrete iff every singleton set is open. If M is a finite metric space and $x\in M$. Let $\epsilon$ be the minimum distance from x to  other points of M, the $B_{\epsilon}(x)$ contains x only So $\{x\}$ is open for every x.So M is discrete.
A: X is a finite metric space and A is a proper subset of X. Take A = {$x_1,x_2,....x_n$}. If A is finite then X\A is finite.Let $x_i\in A$ and one can choose r = min {$d(x_i,x_j)$ | $j \not= i$}. Observe that $B(x_i,r)=${x}$\subset A $. This means every subset of X is open  implies A is open. Also compliment of A  i.e. X\A is also open as X\A is finite. Thus A is closed and open. ie. $\overline A =A $. Hence $\overline A =X $. this implies that A=X as  $\overline X =X $. Thus only dense subset of X is itself.   
A: Hint: If $(A,d)$ is a finite metric space and $x \in A$ and we let
$$\delta=\min_{y \in A \setminus \{x\}}d(x,y)$$
then what is in $B(x,\delta)$?
A: Let $A$ be a finite metric space thought of as a finite set equipped with a metric $\mu \mathop{:} A \to \mathbb{R}_{\ge 0}$ .
Let $\tau$ denote the set of open sets on $A$ induced by the metric $\mu$ .
$\tau$ is generated by a basis $B$ (2), constructed by considering $\varepsilon$ balls of every possible size, here is an explicit formula for $B$ , defined in terms of $B_\alpha$ the set of balls in $A$ centered at $\alpha$.
$$ B \stackrel{\text{def}}{=} \bigcup_{\alpha \in A} B_\alpha \tag{1a} $$
$$ B_{\alpha} \stackrel{\text{def}}{=} \bigcup_{\varepsilon \in \mathbb{R}} \left\{ x \text{ where } x \in A \text{ and } \mu(x,\alpha) < \varepsilon \right\} \tag{1b} $$
$B$ generates $\tau$ by definition since $\tau$ is the topology induced by the metric $\mu$.
$$ \tau \stackrel{\text{def}}{=} \langle B \rangle \tag{2} $$
We want to show that every subset of $A$ is in $\tau$ and hence is open.
Let $\mathcal{F}$ denote the subsets of $A$, specifically $\mathcal{P}(A)$ .
Let $f$ denote an arbitrary element of $\mathcal{F}$ .
Suppose $f$ is empty. Then $f \in \tau$ by the fact that $\tau$ is a topology.
Suppose $f = \left\{z\right\}$ is a singleton. Let $\lambda \in \mathbb{R}_{\ge 0}$ denote the smallest nonzero distance between two points in $A$; $\lambda$ is well-defined because $A$ is finite. Consider the epsilon ball at $z$ with radius $\frac{\lambda}{2}$ ; this ball is a singleton containing only $z$ . $f \in \mathbb{B}$ therefore $f \in \tau$ .
Suppose $f$ is non-empty and isn't a singleton. $f$ is a set, therefore $f$ is a union of singleton sets. Every singleton set in $A$ is in $\tau$ by our earlier argument, therefore $f$ is a union of elements of $\tau$ . Therefore $f$ is in $\tau$, by the fact that $\tau$ is a topology.
A: Metric spaces are $T_1$.  Thus points are closed.   Then, as always,  finite unions of closed sets are closed.  Thus, if the space is finite, all the sets are closed.  In particular,  the complements of all the sets are closed.   Thus the sets are all open.  (So $A$ has the discrete topology.) 
