Definition:Convergence

Convergent Sequence

Topological Space

Let $T = \left({A, \vartheta}\right)$ be a topological space.

Let $\left \langle {x_k} \right \rangle$ be a sequence in $T$.

Then $\left \langle {x_k} \right \rangle$ converges to the limit $\alpha \in T$ if:

for any open set $U \in \vartheta$ such that $\alpha \in U$: $\exists N \in \R: n > N \implies x_n \in U$

This can be alternatively stated:

$\left \langle {x_k} \right \rangle$ converges to the limit $\alpha \in T$ if:

every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\left \langle {x_n} \right \rangle$.

Such a sequence is convergent.

Metric Space

Let $\left({X, d}\right)$ be a metric space.

Let $\left \langle {x_k} \right \rangle$ be a sequence in $X$.

Then $\left \langle {x_k} \right \rangle$ converges to the limit $l$ iff:

$\forall \epsilon > 0: \exists N \in \R: n > N \implies d \left({x_n, l}\right) < \epsilon$

Or equivalently, using the definition of neighborhood:

$\forall \epsilon > 0: \exists N \in \R: n > N \implies x_n \in N_\epsilon \left({l}\right)$

We can write:

$x_n \to l$ as $n \to \infty$

or:

$\displaystyle \lim_{n \to \infty} x_n \to l$

This is voiced:

As $n$ tends to infinity, $x_n$ tends to (the limit) $l$.

It can be seen that by the definition of open set in a metric space, this definition is equivalent to that for convergence in a topological space.

Comment

The sequence $x_1, x_2, x_3, \ldots, x_n, \ldots$ can be thought of as a set of approximations to $l$, in which the higher the $n$ the better the approximation.

The distance $\left|{x_n - l}\right|$ between $x_n$ and $l$ can then be thought of as the error arising from approximating $l$ by $x_n$.

Note the way the definition is constructed.

Given any value of $\epsilon$, however small, we can always find a value of $N$ such that ...

If you pick a smaller value of $\epsilon$, then (in general) you would have to pick a larger value of $N$ - but the implication is that, if the sequence is convergent, you will always be able to do this.

Note also that $N$ depends on $\epsilon$. That is, for each value of $\epsilon$ we (probably) need to use a different value of $N$.

Standard Number Fields

When $X$ is one of the standard number fields $\Q, \R, \C$, and the metric $d$ is the usual (Euclidean) metric, the condition on convergence becomes:

$\left \langle {x_k} \right \rangle$ converges to the limit $l$ iff:

$\forall \epsilon > 0: \exists N \in X: n > N \implies \left|{x_n - l}\right| < \epsilon$

where $\left|{x}\right|$ is the modulus of $x$.

The validity of this definition derives from the fact that:

• the Rational Numbers form Metric Space
• the Real Number Line is Metric Space
• the Complex Plane is Metric Space.

Convergent Series

Let $\left({S, \circ, \tau}\right)$ be a topological semigroup.

Let $\displaystyle \sum_{n=1}^\infty a_n$ be a series in $S$.

This series is said to be convergent iff its sequence $\left \langle {s_N} \right \rangle$ of partial products converges in the topological space $\left({S, \tau}\right)$.

If $s_N \to s$ as $N \to \infty$, the series converges to the sum $s$, and one writes $\displaystyle \sum_{n=1}^\infty a_n = s$.

Convergent Series in a Normed Vector Space

Let $V$ be a normed vector space.

Let $d$ be the induced metric on $V$.

Let $\displaystyle \sum_{n=1}^\infty a_n$ be a series in $V$.

This series is said to be convergent iff its sequence $\left \langle {s_N} \right \rangle$ of partial sums converges in the metric space $\left({V, d}\right)$.

Convergent Series in a Number Field

Let $S$ be one of the standard number fields $\Q, \R, \C$.

Let $\displaystyle \sum_{n=1}^\infty a_n$ be a series in $S$.

Let $\left \langle {s_N} \right \rangle$ be the sequence of partial sums of $\displaystyle \sum_{n=1}^\infty a_n$.

It follows that $\left \langle {s_N} \right \rangle$ can be treated as a sequence in the metric space $S$.

If $s_N \to s$ as $N \to \infty$, the series converges to the sum $s$, and one writes $\displaystyle \sum_{n=1}^\infty a_n = s$.

A series is said to be convergent if it converges to some $s$.

Convergent Function

Convergence of a Function on a Metric Space

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 $f \left({x}\right)$ tend to the limit $L$ as $x$ tends to $c$.

Then $f$ converges to the limit $L$ as $x$ tends to $c$.

Convergence of Real and Complex Functions

As:

• The real number line $\R$ under the usual metric forms a metric space;
• The complex plane $\C$ under the usual metric forms a metric space;

the definition holds for real and complex functions.

Convergent Filter

Let $\left({X, \vartheta}\right)$ be a topological space.

Let $\mathcal F$ be a filter on $X$.

Then $\mathcal F$ converges to a point $x \in X$ if:

$\forall N_x \subseteq X: N_x \in \mathcal F$

where $N_x$ is a neighborhood of $x$.

That is, a filter is convergent to a point $x$ if every neighborhood of $x$ is an element of that filter.

Convergent Filter Basis

Let $\left({X, \vartheta}\right)$ be a topological space.

Let $\mathcal B$ be a filter basis of a filter $\mathcal F$ on $X$.

Then $\mathcal B$ converges to a point $x \in X$ iff:

$\forall N_x \subseteq X: \exists B \in \mathcal B: B \subseteq N_x$

where $N_x$ is a neighborhood of $x$.

That is, a filter basis is convergent to a point $x$ if every neighborhood of $x$ contains some set of that filter basis.