Metric Space Continuity by Epsilon-Delta

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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:

$\epsilon$-$\delta$ Definition

$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) if and only if:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$

where $\R_{>0}$ denotes the set of all strictly positive real numbers.

Definition by Limits

$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) if and only if:

$(1): \quad$ The limit of $f \left({x}\right)$ as $x \to a$ exists
$(2): \quad \displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$.


Proof

Definition by Limits implies $\epsilon$-$\delta$ Definition

Suppose that $f$ is $\left({d_1, d_2}\right)$-continuous at $a$ in the sense that:

$(1): \quad$ The limit of $f \left({x}\right)$ as $x \to a$ exists
$(2): \quad \displaystyle \lim_{x \to a} f \left({x}\right) = f \left({a}\right)$.


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

Then by the $\epsilon$-$\delta$ definition of limit:

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

By $(2)$:

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

that is:

$f \left({a}\right) = L$

Then:

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

and so:

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

Thus the definition by limits implies the $\epsilon$-$\delta$ definition. {{qed|lemma}


$\epsilon$-$\delta$ Definition implies Definition by Limits

Suppose that $f$ is $\left({d_1, d_2}\right)$-continuous at $a$ in the sense that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon$

where $\R_{>0}$ denotes the set of all strictly positive real numbers.

There are two possibilities for $d_1 \left({x, a}\right)$:

$(A): \quad 0 < d_1 \left({x, a}\right) \delta$

Then by the $\epsilon$-$\delta$ definition of limit:

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

$(B): \quad d_1 \left({x, a}\right) = 0$

That is, $x = a$.

Hence:

$f \left({x}\right) = f \left({a}\right)$

and so again:

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


Thus the $\epsilon$-$\delta$ definition implies the definition by limits.

$\blacksquare$


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