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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 $\epsilon$?

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It says that we can choose one such $x$, not necessarily all $x$ must satisfy it. –  awllower Feb 12 '13 at 6:53
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$f$ is continuous on $M$ if and only if $$ \forall p\in M\,\forall\varepsilon>0\,\exists\delta>0\,\forall x\in M:d_N(x,p)<\delta\Rightarrow d_M(f(x),f(p))<\varepsilon $$ so $f$ is not continuous on $M$ if the contrapositive holds, that is $$ \exists p\in M\,\exists\varepsilon>0\,\forall \delta>0\,\exists x\in M: d_N(x,p)<\delta\wedge d_M(f(x),f(p))\geq \varepsilon. $$ Now assume $f$ is not continuous, and let $p\in M$ and $\varepsilon>0$ be the $p$ and $\varepsilon$ that we know exist from above. Then for any $\delta>0$ in particular for $\delta_n=\frac{1}{n}$ we know that there exists an $x\in M$ (call this $x_n$) such that $$ d_N(x_n,p)<\delta \wedge d_M(f(x_n),f(p))\geq \varepsilon. $$

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