# Equivalence of Definitions of Limit of Function in Metric Space

## Theorem

The following definitions of the concept of Limit of Function in the context of Metric Spaces are equivalent:

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $c$ be a limit point of $M_1$.

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$ defined everywhere on $A_1$ except possibly at $c$.

Let $L \in M_2$.

### $\epsilon$-$\delta$ Condition

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < d_1 \left({x, c}\right) < \delta \implies d_2 \left({f \left({x}\right), L}\right) < \epsilon$

That is, for every real positive $\epsilon$ there exists a real positive $\delta$ such that every point in the domain of $f$ within $\delta$ of $c$ has an image within $\epsilon$ of some point $L$ in the codomain of $f$.

### $\epsilon$-Ball Condition

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$.

where:

$B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}$ is the deleted $\delta$-neighborhood of $c$ in $M_1$
$B_\epsilon \left({L; d_2}\right)$ is the open $\epsilon$-ball of $L$ in $M_2$.

That is, for every open $\epsilon$-ball of $L$ in $M_2$, there exists a deleted $\delta$-neighborhood of $c$ in $M_1$ whose image is a subset of that open $\epsilon$-ball.

## Proof

### $\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition

Suppose that $f$ satisfies the $\epsilon$-$\delta$ condition:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < d_1 \left({x, c}\right) < \delta \implies d_2 \left({f \left({x}\right), L}\right) < \epsilon$

Let $y \in f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right)$.

By definition of open $\epsilon$-ball, this means:

$\exists x \in A_1: 0 < d_1 \left({x, c}\right) < \delta: y = f \left({x}\right)$

By hypothesis, it follows that $d_2 \left({y, L}\right) < \epsilon$

That is, that $y \in B_\epsilon \left({L; d_2}\right)$.

By definition of subset:

$f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$

Thus it follows that $f$ satisfies the $\epsilon$-ball condition.

$\Box$

### $\epsilon$-Ball Condition implies $\epsilon$-$\delta$ Condition

Suppose that $f$ satisfies the $\epsilon$-ball condition:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$.

Let $0 < d_1 \left({x, c}\right) < \delta$.

Then by definition of open $\epsilon$-ball, this means:

$x \in B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}$

By hypothesis, it follows that:

$f \left({x}\right) \in B_\epsilon \left({L; d_2}\right)$

Thus by definition of open $\epsilon$-ball, this means:

$d_2 \left({f \left({x}\right), L}\right) < \epsilon$

That is, $f$ satisfies the $\epsilon$-$\delta$ condition.

$\blacksquare$