Definition:Convergent Sequence (Metric Spaces)

Definition

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 $N$.

Note

Some sources insist that $N \in \N$ but this is not strictly necessary and can make proofs more cumbersome.