Let \((X,d)\) be a metric space and \((x_n)\) a sequence on \(X\). The sequence converges to some \(x\in X\) if and only if for all \(\varepsilon>0\) there is a \(n_0\in \mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) for all \(n\ge n_0\).
Proof
Let \(\varepsilon>0\). Since the ball \(B_\varepsilon(x)\) is open there is a number \(n_0\), such that \(x_n\in B_{\varepsilon}(x)\) for all \(n\ge n_0\).
On the contrary, consider some open neighbourhood \(U\) of \(x\). Then there is an open ball \(B_\varepsilon(x)\subseteq U\). So there is a \(n_0\in \mathbb{N}\) such that \(x_n\in B_\varepsilon(x)\subseteq U\) for all \(n\ge n_0\).