So, assuming whoever can answer this knows the first part of this proof, that is, showing that for all$ \epsilon\gt0$ and for all $ p\in M$ there is a $\delta\gt0$. Also there is an $ x\in M$ such that $d_M(x,p)\lt\delta\implies d_N(f(x),f(p))\lt \epsilon$. I may not have this perfectly right but I'd assume you can understand. Now I get and understand this. But what I don't understand is the showing that the contrapositive leads to a contradiction. So my book (using Real Analysis by Charles Pugh) says this as the proof:
If $f$ is not continuous then it does not preserve convergence. $f$ being not continuous means that for some $p\in M$ there is an $\epsilon\gt 0$ such that no matter how small we take $\delta$, there will always be points $x\in M$ with $d(x,p)\lt \delta$ but $d(fx,fp)\ge \epsilon$. Take $\delta_1=1, \delta_2=\frac12,... \delta_n=\frac1n$. For each $\delta_n$ there is a point $x_n$ with $d(x_n,p)\lt \delta_n=\frac1n$ and $d(f(x_n),fp)\ge \epsilon$. Thus $$\lim_{n\to \infty}x_n=p$$ but $f(x_n)$ does not converge to $fp$.
My main confusion is why the fact that $d(x,p)\lt \delta \implies d(fx,fp)\ge \epsilon$. Why is it that just because we decide to make $\delta$ really small then $d(fx,fp)$ is all of a sudden bigger than $\epsilon$?
