1. ## completeness

Let $(M_1,d_1)$ and $(M_2,d_2)$ be metric spaces and let $f:M_1 \rightarrow M_2$ be a continuous surjective map such that $d_1(p,q) \leq d_2(p,q)$ for evey $p,q \in M_1$.
a)If $M_1$ is complete, is $M_2$ complete?
b)If $M_2$ is complete, is $M_1$ complete?

2. Kat-M
Will you please tell us about the source of the problems that you have posted?
So far as I can see, you have made no effort at any solutions whatsoever.
Are these from a list of prelims for your degree program?
These questions are so wide ranging, I must ask you to explain your question

3. ## questions

yes i am studying for the prelim and these problems are from old prelims like 5 to 10 years ago. i am trying to get help from a lot of people to understand these problems. and also i am trying to solve them myself. like this one i posted , i think i got the second part.

Assume that $M_2$ is complete.
Let ${x_n}$ be Caushy in $M_1$.And using the fact that $f$ is const, for $\epsilon \geq 0$ there exists $N$ such that $n,m \geq N$ implies $d_1(x_n,x_m)< \delta$ such that $d_2(f(x_n),f(x_m)) < \epsilon$. so $\{f(x_n)\}$ is also Caushy in $M_2$.
Since $M_2$ is complete, $f(x_n) \rightarrow y \in M_2$. Since f is surjective, there exists $x \in M_1$ such that $f(x)=y$.
Since $\{f(x_n)\}$ converges to $y$, for $\epsilon > 0$, there exists $M$ such that $n \geq M$ implies $d_2(f(x_n),y) < \epsilon$. Since $d_1 \leq d_2$, $d(x_n, x) < \epsilon$. So $\{x_n\}$ converges to $x$. Thus $(M_1, d_1)$ is complete.

I am stuck in part 1 so if you can help me with this, i am really greatful. Thank you.