(related to Proposition: (\epsilon)-(\delta) Definition of Continuity)
We have to show that for two given metric spaces \((X,d_x)\) and \((Y,d_y)\) and a function \(f:X\to Y\) the definition of continuity of \(f\) at a point \(a\in X\) is equivalent to the \(\epsilon\)-\(\delta\) definition of continuity.
Let \(f\) be continuous at \(a\), which, according to first definition means that \(\lim_{x\to a} f(x)=f(a)\). Assume, the \(\epsilon\)-\(\delta\) condition was not fulfilled. Then there would be an \(\epsilon > 0\) such that for every \(\delta > 0\) we would find a point \(x\in X\) such that
\[d_x(x,a) < \delta\]
but
\[d_y(f(x),f(a)) \ge \epsilon.\]
In particular, for a \(\delta:=\frac 1n\) we would have a sequence of points \(x_n\in X\) such that \[d_x(x_n,a) < \frac 1n\quad\wedge\quad d_y(f(x_n),f(a)) \ge \epsilon\quad\quad ( * ) \]
However, the left inequation means that \(\lim x_n=a\) and so \(\lim f(x_n)=f(a)\), in contradiction to \(( * )\). Therefore, it must be that the \(\epsilon\)-\(\delta\) condition is fulfilled.
Let \((x_n)_{n\in\mathbb N}\) be a sequence of points with \(\lim x_n=a\). We have to show that \(\lim f(x_n)=f(a)\), when the \(\epsilon\)-\(\delta\) condition be fulfilled.
From \(\lim x_n=a\) it follows that for every (arbitrarily small,) given \(\delta > 0\) there exists \(N(\delta)\in\mathbb N\) such that \(d_x(x_n,a) < \delta\) for all \(n \ge N(\delta)\). Now, since the \(\epsilon\)-\(\delta\) condition is fulfilled, we have that for every given \(\epsilon > 0\) there is a \(\delta > 0\) such that from \[d_x(x_n,a) < \delta.\] it follows that \[d_y(f(x_n),f(a)) < \epsilon.\] Because this is true for all \(n \ge N(\epsilon):=N(\delta)\), it is equivalent to \(\lim f(x_n)=f(a)\).
