Prove wrong the following statement about metric spaces and completeness

Question

Statement:

Given the condition:

$d(x,y)^2 \leq g(x,y) \leq d(x,y) \space \forall \space x,y \in M$

If $(M,d)$ is complete then $(M,g)$ is complete

Question:

Prove or provide a counterexample to the above statement.

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I am having serious difficulty coming up with 2 distance functions which satisfy the above condition, let alone finding a counterexample.

I've considered the distance functions

$d(x,y)=\left\{\begin{matrix} 0 \iff x=y\\ \frac{1}{2} \iff x \neq y \end{matrix}\right.$

and

$g(x,y)=\left\{\begin{matrix} 0 \iff x=y\\ \frac{1}{3} \iff x \neq y \end{matrix}\right.$

These satisfy the condition but both are complete in any $M$ assuming $\frac{1}{2}, \frac{1}{3} \in M$

Can someone explain an intuitive and methodical approach to finding such an answer. Even walking me to the answer without giving me it explicitly is fine. I'd just like to know how I might approach it. Thanks.

Answer 1

$(x_n)$ Cauchy in $(M,g)\implies g(x_n,x_m)\to 0 $ as $n,m\to \infty\implies $

$d(x_n,x_m)^2\leq g(x_n,x_m)\to 0 \implies d(x_n,x_m)\to 0\implies x_n$ is a Cauchy in $(M,d)$

$\implies x_n\to x$ for some $x\in M\implies d(x_n,x)\to 0$ as $n\to \infty$

$\implies g(x_n,x)\leq d(x_n,x)\to 0\implies x_n\to x$ in $(M,g)$

Thus $(M,g)$ is complete

Answer 2

There is no counterexample because the statement is true. To show that this is the case: first, show that a sequence is Cauchy in one metric iff it is Cauchy in the other. Next, show that $x_n\to x$ in one metric iff $x_n\to x$ in the other.