# Every finite set of any metric space is complete

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. $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Box$$

• Do you have a question to ask? Commented Oct 8, 2019 at 4:14
• Yes, proof verification. This is my attempt at proving the above question and I would like to see if my proof is correct hence, the "proof-verification" tag.
– Sun
Commented Oct 8, 2019 at 4:15
• Your proof is correct, so it's hard for people to post an answer - that would be just "Correct". Commented Oct 8, 2019 at 4:31

## 1 Answer

• $$d$$ 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.

• And the very first sentence lacks a word "subset". Commented Oct 8, 2019 at 4:32
• Thank you! As I was writing down my proof I realized the very first point and I made sure to write $x \neq y$. I greatly appreciate the personal remark.
– Sun
Commented Oct 8, 2019 at 4:35
• Ah! Grammar. Thanks.
– Sun
Commented Oct 8, 2019 at 4:45
• @Sunny If I point out a mistake in your post after you've asked me to do so, then you should not change it. It makes the answer look silly, and in truth people should not be bothered with reviewing their answers over and over. It doesn't matter right now, but keep it in mind.
– user239203
Commented Oct 8, 2019 at 4:56
• Okay makes sense.
– Sun
Commented Oct 8, 2019 at 11:35