# Proof

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.

### "$$\Rightarrow$$"

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.

### "$$\Leftarrow$$"

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 $$\lim f(x_n)=f(a)$$.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984