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 \(d(x_n, y_n) \to d(x,y)\), \((n \to \infty)\).
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
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
by axiom 4 of the definition of a metric space. Hence we obtain
such that
as \(n \to \infty\).