Proving $(X, d)$ is complete if and only if $(X, d')$ is complete

Question

My question is how to prove that $(X,d)$ is complete if and only if $(X,d')$ is complete.

I have that $d$ and $d'$ are strongly equivalent metrics and I have used this to show that a sequence $x_{n}$ is Cauchy in $(X,d)$ if and only if it is Cauchy in $(X,d')$.

I have the definition of complete as: "A metric space $X$ is complete if every Cauchy sequence in $X$ is convergent in $X$."

Since this is an if and only if statement I know I need to prove it both ways.

I am wondering if you also use the definition of strongly equivalent metrics to prove the completeness of the metrics or if you need to prove the convergence of the Cauchy sequence. A bit confused on how to approach this.

Answer 1

So $\alpha d'(x,y)\leq d(x,y)\leq\beta d'(x,y)$ for some $\alpha,\beta>0$.

Given $d(x_{n},x_{m})\rightarrow 0$ and $d'$ is complete, then $\alpha d'(x_{n},x_{m})\rightarrow 0$, so $d'(x_{n},x_{m})\rightarrow 0$, then for some $x\in X$, $d'(x_{n},x)\rightarrow 0$, and hence $\dfrac{1}{\beta}d(x_{n},x)\rightarrow 0$, and so $d(x_{n},x)\rightarrow 0$.

Answer 2

Hint: You are assuming that $d$ and $d'$ are strongly equivalent, right?! Now, given a sequence $(x_n)_{n\in\mathbb N}$ of elements of $X$, prove that

1. if $x\in X$, then $(x_n)_{n\in\mathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,d')$;
2. $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence in $(X,d)$ if and only if it is a Cauchy sequence in $(X,d')$.