1.8 KiB
Convergence
Definition 1: a sequence
(x_n)_{n \in \mathbb{N}}
in a metric space(X,d)
is convergent if there exists anx \in X
such that
\lim_{n \to \infty} d(x_n, x) = 0.
x
is the limit of(x_n)
and is denoted by
\lim_{n \to \infty} x_n = x,
or simply by
x_n \to x
,(n \to \infty)
.
We say that (x_n)
converges to x
or has the limit x
. If (x_n)
is not convergent then it is divergent.
We have that the limit of a convergent sequence must be a point of X
.
Definition 2: a non-empty subset
M \subset X
of a metric space(X,d)
is bounded if there exists anx_0 \in X
and anr > 0
such thatM \subset B(x_0,r)
.
Furthermore, we call a sequence (x_n)
in X
a bounded sequence if the corresponding point set is a bounded subset of X
.
Lemma 1: let
(X,d)
be a metric space then
- a convergent sequence in
X
is bounded and its limit is unique,- if
x_n \to x
andy_n \to y
thend(x_n, y_n) \to d(x,y)
,(n \to \infty)
.
??? note "Proof:"
For statement 1, suppose that $x_n \to x$. Then, taking $\varepsilon = 1$, we can find an $N$ such that $d(x_n, x) < 1$ for all $n > N$. Which shows that $(x_n)$ is bounded. Suppose that $x_n \to x$ and $x_n \to z$ then by axiom 4 of the definition of a metric space we have
$$
d(x_n, x) \leq d(x_n, z) + d(x, z) \to 0,
$$
as $n \to \infty$ and by axiom 2 of the definition of a metric space it follows that $x = z$.
For statement 2, we have that
$$
d(x_n,y_n) \leq d(x_n, x) + d(x, y) + d(y, y_n),
$$
by axiom 4 of the definition of a metric space. Hence we obtain
$$
d(x_n, y_n) - d(x, y) \leq d(x_n, x) + d(y_n, y),
$$
such that
$$
|d(x_n, y_n) - d(x, y)| \leq d(x_n, x) + d(y_n, y) \to 0
$$
as $n \to \infty$.