Theorem

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

Let $S \subseteq A_1$ be an open set of $M_1$.

Let $f$ be a mapping defined on $S$, except possibly at the point $c \in S$.



Then $\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$ iff:

for each sequence $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\forall n \in \N_{>0}: x_n

e c$ and $\displaystyle \lim_{n \to \infty} x_n = c$

it is true that:

$\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$





On the real number line, this result becomes as follows:



Let $f$ be a real function defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.



Then $\displaystyle \lim_{x \mathop \to c} \map f x = l$ if and only if:

for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n

e c$ and $\displaystyle \lim_{n \to \mathop \infty} x_n = c$

it is true that:

$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$





The above result holds for a real function tending to a limit both from the right and from the left:



Let $\openint a b$ be an open real interval.

Let $f: \openint a b \to \R$ be a real function.

Let $l \in \R$.



Then:

$(1): \quad \displaystyle \lim_{x \mathop \to a^+} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x_n = a \implies \lim_{n \mathop \to \infty} \map f {x_n} = l$ $(2): \quad \displaystyle \lim_{x \mathop \to b^-} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x_n = b \implies \lim_{n \mathop \to \infty} \map f {x_n} = l$

where:





Proof

Necessary Condition

Suppose that:

$\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$

Let $\epsilon > 0$.

Then by the definition of the limit of a function:

$\exists \delta > 0: d_2 \left({f \left({x}\right), l}\right) < \epsilon$

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

Now suppose that $\left \langle {x_n} \right \rangle$ is a sequence of points of $S$ such that such that:

$\forall n \in \N_{>0}: x_n

e c$ and $\displaystyle \lim_{n \to \infty} x_n = c$

Since $\delta > 0$, from the definition of the limit of a sequence:

$\exists N: \forall n > N: d_1 \left({x_n, c}\right) < \delta$

But:

$\forall n \in \N_{>0}: x_n

e c$

That means:

$0 < d_1 \left({x_n, c}\right) < \delta$

by the definition of a metric.

But that implies that:

$d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$

That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:

$\forall n > N: d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$

Thus:

$\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$

$\Box$





Sufficient Condition

Now suppose that for each sequence $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\displaystyle \forall n \in \N_{>0}: x_n

e c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, it is true that:

$\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$

What we will try to do is assume that it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, and try to find a contradiction.

So, if it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, then:

$\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < d_1 \left({x, c}\right) < \delta \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon$

where $\land$ denotes logical and.

For all $n \in \N_{>0}$, define:

$S_n = \left\{{x \in S: 0 < d_1 \left({x, c}\right) < \dfrac 1 n \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon}\right\}$

By hypothesis, $S_n$ is non-empty for all $n \in \N_{>0}$.

Using the axiom of countable choice, there exists a sequence $\left \langle {x_n} \right \rangle$ of points in $S$ such that $x_n \in S_n$ for all $n \in \N_{>0}$.

Then:

$\forall n \in \N_{>0}: x_n

e c$ and $\displaystyle \lim_{n \to \infty} x_n = c$

but it is not true that:

$\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$

So there is our contradiction, and so the result follows.

$\blacksquare$





Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.



