Question in Construction for Proof of Nested Sphere Theorem

Question

Not Homework Just Personal Study:

Statement of Theorem: A metric space $(M,d)$ is complete if and only if every nested sequence of closed balls $S_n=\{B(x_n,r_n)\}$ with $\lim_{n\to \infty}r_n=0$ satisfies $\cap_{n=1}^{\infty}S_n\neq \emptyset.$

The forward direction is clear. For the backward direction, the following construction is given.

Let $x_n$ be cauchy in $(M,d).$ We want to show that $x_n$ converges to $x \in M.$ By that fact that $x_n$ is cauchy, for each $i \in \mathbb{N}$ we can pick $x_{n_i}$ such that $d(x_n,x_{n_i})<\frac{1}{2^i}.$ Let $S_{k+i}$ be the closed sphere of radius $\frac{1}{2^{k+i-1}}$ with center $x_{n_{k+i}}.$ I do not understand why the $S_{k+i}$ are nested. We want to show that $S_{k+2} \subseteq S_{k+1}$ for all $k.$ If $y \in S_{k+2}$ then we know that $d(x_n,x_{n_{k+i}})<\frac{1}{2^{k+i}}$ and we know that $d(x_{n_{k+2}},y)<\frac{1}{2^{k+2}}.$ Then

$$ d(x_{n_{k+1}},y) \leq d(x_{n_{k+1}},x_{n_{k+2}})+d(x_{n_{k+2}},y)\leq d(x_{n_{k+1}},x_n)+d(x_{n_{k+2}},x_n)+d(x_{n_{k+2}},y)$$

So that

$$d(x_{n_{k+1}},y)\leq \frac{1}{2^{k+1}}+\frac{1}{2^{k+2}}+\frac{1}{2^{k+1}} =\frac{1}{2^k}+\frac{1}{2^{k+2}}$$

But we need to show that $d(x_{n_{k+1}},y)\leq \frac{1}{2^k}$

I'm not sure why this is true or what I am missing.

Thank you in advanced for any help.

Answer 1

The construction you give needs clarification, and the lack of clarity is likely leading to your confusion. When you write

By that fact that $x_n$ is cauchy, for each $i\in \mathbb N$ we can pick $x_{n_i}$ such that $d(x_n,x_{n_i})<\frac{1}{2^i}$,

you (or whoever gave the argument you are citing) need to quantify that this is true for all $n>n_i$. And at later points in the argument, you also refer to $x_n$ without specifying which $n$ you are talking about. This turns out to be a source of confusion for you later in the proof.

Also, the definition of $S_{k+i}$ really doesn't need two indices - you could just as easily define $S_k$ to be the closed ball of radius $\frac{1}{2^{k-1}}$ centered at $x_{n_k}$. It's not strictly speaking incorrect, just very, very awkward and confusing notation.

Now, having clarified that, in the line

$$d(x_{n_{k+1}},y) \leq d(x_{n_{k+1}},x_{n_{k+2}})+d(x_{n_{k+2}},y)\leq d(x_{n_{k+1}},x_n)+d(x_{n_{k+2}},x_n)+d(x_{n_{k+2}},y)\text{,}$$

the final inequality is completely unnecessary, and you can simply directly write

$$d(x_{n_{k+1}},y) \leq d(x_{n_{k+1}},x_{n_{k+2}})+d(x_{n_{k+2}},y)\leq \frac{1}{2^{k+1}} + \frac{1}{2^{k+1}}=\frac{1}{2^k}\text{.}$$

The second inequality follows from the definition of $S_{k+2}$, as well as the earlier remark that $d(x_n,x_{n_i})<\frac{1}{2^i}$ holds for all $n>n_i$, including in particular, $n=n_{i+1}$.