Contraposition of continuity Function f is continuous on a metric space (M,d) if
$$\forall x\ \forall \epsilon >0\ \exists \delta >0\ \forall y\ \ d(x,y)< \delta \Rightarrow d(f(x),f(y))<\epsilon $$
I want to find a contrapositive of this. I know that I have to change $\exists$ to $\forall$ and $\forall$ to $\exists$ but I am confused on what happens with " $\Rightarrow$".
$$\exists x\ \exists\epsilon>0\ \forall \delta>0\ \exists y\ \text{ such that... } $$ 
And then I am stuck. Would it be $d(f(x),f(y))\geq\epsilon \text{ but } d(x,y) \geq \delta$ ?

The contraposition of the implication is: we have $y \in M$ and $d(f(x),f(y)) \geq \varepsilon$ only if $d(x,y) \geq \delta$.
The negation of the whole statement is: there are some $\varepsilon > 0$ and some $x \in M$ such that $\delta > 0$ only if there is some $y \in M$ such that $d(x,y) < \delta$ and $d(f(x), f(y)) \geq \varepsilon$.
