[proofplan] We prove uniqueness by contradiction. If two proposed limits $x$ and $y$ were distinct, the [Hausdorff property](/page/Hausdorff%20Space) would provide disjoint [open neighbourhoods](/page/Open%20Set) of them. Convergence to each point then forces the tail of the same [sequence](/page/Sequence) to lie in both disjoint open sets, which is impossible. The [metric-space](/page/Metric%20Space) statement follows by applying the triangle inequality directly. [/proofplan] [step:Separate two distinct proposed limits by disjoint open neighbourhoods] Assume, toward a contradiction, that $x \neq y$. Since $(X,\tau)$ is a Hausdorff space, there exist sets $U, V \in \tau$ such that $x \in U$, $y \in V$, and $U \cap V = \varnothing$. [/step] [step:Force one tail of the sequence into both neighbourhoods] Since $x_n \to x$ in $(X,\tau)$ and $U \in \tau$ is an open set containing $x$, the definition of convergence in a [topological space](/page/Topological%20Space) gives an index $N_1 \in \mathbb{N}$ such that $x_n \in U$ for every $n \geq N_1$. Since $x_n \to y$ in $(X,\tau)$ and $V \in \tau$ is an open set containing $y$, the same definition gives an index $N_2 \in \mathbb{N}$ such that $x_n \in V$ for every $n \geq N_2$. Define $N := \max\{N_1,N_2\}$. Then $N \geq N_1$ and $N \geq N_2$, so $x_N \in U$ and $x_N \in V$. Hence $x_N \in U \cap V$, contradicting $U \cap V = \varnothing$. Therefore the assumption $x \neq y$ is false, and $x = y$. [guided] We use the definition of convergence in a topological space twice, once for each alleged limit. Because $U$ is an open set containing $x$ and $x_n \to x$, there is an index $N_1 \in \mathbb{N}$ such that every term from that point onward lies in $U$: \begin{align*} n \geq N_1 \implies x_n \in U. \end{align*} Likewise, because $V$ is an open set containing $y$ and $x_n \to y$, there is an index $N_2 \in \mathbb{N}$ such that \begin{align*} n \geq N_2 \implies x_n \in V. \end{align*} The purpose of taking a maximum is to find a single index after both eventual behaviours have started. Define $N := \max\{N_1,N_2\}$. Then $N \geq N_1$ and $N \geq N_2$, so the two implications give both $x_N \in U$ and $x_N \in V$. Therefore $x_N \in U \cap V$. But the Hausdorff separation step gave $U \cap V = \varnothing$, so no such point $x_N$ can exist. This contradiction proves that the assumption $x \neq y$ was impossible, and hence $x = y$. [/guided] [/step] [step:Prove the metric-space statement by the triangle inequality] Let $(X,d)$ be a metric space, and let $\{x_n\}_{n=1}^{\infty}$ be a sequence in $X$ satisfying $d(x_n,x) \to 0$ and $d(x_n,y) \to 0$. Assume, toward a contradiction, that $x \neq y$. By the separation axiom for a metric, $d(x,y) > 0$. Define $r := d(x,y)/2$. Since $d(x_n,x) \to 0$, there exists $N_1 \in \mathbb{N}$ such that $d(x_n,x) < r$ for every $n \geq N_1$. Since $d(x_n,y) \to 0$, there exists $N_2 \in \mathbb{N}$ such that $d(x_n,y) < r$ for every $n \geq N_2$. Define $N := \max\{N_1,N_2\}$. By the triangle inequality, \begin{align*} d(x,y) \leq d(x,x_N) + d(x_N,y) < r + r = d(x,y), \end{align*} which is impossible. Therefore $x = y$. [/step]