Convergence

Definition 1: a sequence (x_n)_{n \in \mathbb{N}} in a metric space (X,d) is convergent if there exists an x \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 an x_0 \in X and an r > 0 such that M \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 and y_n \to y then d(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 $n \to \infty$.

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Convergence

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