Every finite set of any metric space is complete

Question

Let $\left< M, \rho \right>$ be a metric space and let $A \subset M$ be a finite subset.

Let $\{ x_n \}_{n=1}^\infty \in A$ be a Cauchy sequence.

Define $$d= \inf_{\forall x,y \in A, x \neq y} \rho(x,y)$$

Since $\{ x_n \}_{n=1}^\infty$ is Cauchy, $\exists N \in \mathbb{N}$ s.t. if $m,n \geq N$ then $\rho(x_n,x_m) \lt \frac{d}{2}$.

But $x_n, x_m \in A \ \ \ \forall m,n \in \mathbb{N}$ and the minimum distance between any two points in $A$ is $d$.

So $\rho(x_n, x_m) \lt \frac{d}{2} \implies x_n=x_m \forall n,m \geq N$

Or, $x_n = x_m = x_N = x_{N+1} = ... \to \infty$ if $m,n \geq N$

So $\{ x_n \}_{n=1}^\infty \to x_N$, but $x_N \in \{ x_n \}_{n=1}^\infty \in A$.

So a Cauchy sequence in $A$ converges to a point in $A$ hence, $A$ is complete. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Box$

Answer 1

• $d$ defined as you did would be $0$. On the other hand, $d=\inf\limits_{x,y\in A, \,x\ne y} \rho(x,y)$ is a positive number (perhaps you could want to clarify why).

• As a completely personal remark, notation such as $x_n = x_m = x_N = x_{N+1} = ... \to \infty$ is the kind of thing that I don't want to see. I think that if you mean that $x_n=x_{N+1}$ for all $n>N$, then you should write that, if you mean that the sequence is eventually constant, then you should write that, if you mean that $x_n=x_m=x_N=x_{N+1}=...$ to infinity then you should stop meaning that. That "to infinity" is redundant, if not confusing, and you could just simply write $x_{N+1}=x_{N+2}=x_{N+3}=\cdots$ to the same effect.

• Technically, you proved that the sequence is eventually equal to $x_{N+1}$, not $x_N$.

Other than these details I'd say it's fine.