Equivalence of Definitions of Metric Space Continuity at Point

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

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.

Let $a \in A_1$ be a point in $A_1$.

Then the following definitions of continuity of $f$ at $a$ with respect to $d_1$ and $d_2$ are equivalent:

$(1) \iff (2)$
By definition of $\epsilon$-$\delta$ condition definition:


 * $\displaystyle \lim_{x \to a} f \left({x}\right) = L$

iff:


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

Suppose that:
 * $\displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$

that is:
 * $f \left({a}\right) = L$

Then:
 * $0 = d_1 \left({a, a}\right)$

and so:
 * $d_2 \left({f \left({a}\right), L}\right) = 0 < \epsilon$

Now let:
 * $\forall \epsilon \in \R_{>0}: d_2 \left({f \left({a}\right), L}\right) < \epsilon$

Suppose $d_2 \left({f \left({a}\right), L}\right) = e > 0$.

Then $\exists \epsilon \in \R_{>0}: \epsilon < e: d_2 \left({f \left({a}\right), L}\right) > \epsilon$

From this contradiction it follows that $d_2 \left({f \left({a}\right), L}\right) = 0$.

That is:
 * $\displaystyle f \left({a}\right) = L = \lim_{x \to a} f \left({x}\right)$

Hence it follows that $(1) \iff (2)$.

$(2) \iff (3)$
By the $\epsilon$-ball condition definition:


 * $\displaystyle \lim_{x \to a} f \left({x}\right) = L$

iff:


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

Suppose that:
 * $\displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$

that is:
 * $f \left({a}\right) = L$

Then:
 * $f \left({a}\right) \in B_\epsilon \left({L; d_2}\right)$

Suppose that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({a; d_1}\right)}\right) \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right)$

For all $\delta \in \R_{>0}$, we have that:
 * $a \in B_\delta \left({a; d_1}\right)$

Thus it follows that:
 * $\forall \epsilon \in \R_{>0}: f \left({a}\right) \in B_\epsilon \left({L; d_2}\right)$

Using the same argument as above it follows that $f \left({a}\right) = L$.

That is, $\displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$.

Hence it follows that $(2) \iff (3)$.