If $M$ is complete and $f : (M,d)\to(N,p)$ is continuous, then $f(M)$ is complete? Prove or disprove: If $M$ is complete and $f:(M, d )\to (N, p)$ is continuous, 
then $f(M)$ is complete. 
 A: BROAD HINT: Let $M=N=\Bbb Q$, and let $d$ be a discrete metric on $M$, e.g., $$d(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;.\end{cases}$$ Put the right  non-discrete topology on $N$, and use the identity function $f:M\to N:x\mapsto x$. The discrete metric on $M$ ensures that $f$ is continuous no matter what metric you put on $N$.
It will help to ask yourself what are the Cauchy sequences in $\langle M,d\rangle$.
A: In general this is not true as pointed out in several counter-examples/hints on other posts: I'm going to add something else here. Continuity is just too weak to do the job. What you need from $f$ in order for the completeness of $M$ to be transmitted to $f(M)$ is some control over the sequences in $M$ that map to Cauchy-sequences in $f(M)$.
One sufficient property is $f$ being Bi-Lipschitz. One says that a function $f:(X,d)\to (Y,e)$ is Bi-Lipschitz if there exists $M\geq 1$ so that $\frac{1}{M}d(x,y)\leq e(f(x),f(y))\leq Md(x,y)$ for all $x,y\in X$. It is quite straight-forward to show that such property is indeed enough for $f(M)$ to be complete if $M$ is. And in fact, there doesn't seem to be many properties that you could drop out and make it still work: maybe this is close to necessary condition as well.
Added: For completness sake, I will add a counter-example of a Lipschitz function that fails to do the job. Since $[1,\infty[$ is a closed subset $\mathbb{R}$ it is thus complete and $\frac{1}{x}:[1,\infty[\to]0,1]$, denoted by $f$, is a continuous bijection, yet $]0,1]$ is not complete. Choose e.g. the standard Cauchy-sequence $(\frac{1}{n})_{n=1}^{\infty}$ which does not converge in $]0,1]$.
In addition $f\in C^{1}([1,\infty[)$ and $|f\,'(x)|=|-\frac{1}{x^{2}}|=\frac{1}{x^{2}}\leq 1$ so $f$ is $1$-Lipschitz by the intermediate value theorem. So even a Lipschitz function is not enough for the result! Maybe you can figure out something closer to Bi-Lipschitz (but not quite) that also fails.
A: Let $M=N=\mathbb{R}$ and $f(x)={x \over 1+|x|}$. $f$ is of course continuous, $\mathbb{R}$ is complete but its image $(-1,1)$ isn't.
A: Hint consider both spaces as the real numbers with he standard metric and $f$ to be a continuous function which is a bijection with an open interval, e.g. $\arctan$. 
