Convergent Sequence in Hausdorff Space has Unique Limit

Theorem

Let $T = \left({A, \vartheta}\right)$ be a Hausdorff space.

Let $\left \langle {x_n} \right \rangle$ be a convergent sequence in $T$.

Then $\left \langle {x_n} \right \rangle$ has exactly one limit.

Proof

From the definition of convergent sequence, we have that $\left \langle {x_n} \right \rangle$ converges to at least one limit.

Suppose $\displaystyle \lim_{n \to \infty} x_n = l$ and $\displaystyle \lim_{n \to \infty} x_n = m$ such that $l \ne m$.

As $T$ is Hausdorff, $\exists U \in \vartheta: l \in U$ and $\exists V \in \vartheta: m \in V$ such that $U \cap V = \varnothing$.

Then, from the definition of convergent sequence:

	\(\displaystyle \exists N_U:\)	\(\displaystyle n > N_U\)	\(\implies\)	\(\displaystyle x_n \in U\)	\(\displaystyle \)
	\(\displaystyle \exists N_V:\)	\(\displaystyle n > N_V\)	\(\implies\)	\(\displaystyle x_n \in V\)	\(\displaystyle \)

Taking $N = \max \left\{{N_U, N_V}\right\}$ we then have:

$\exists N: n > N \implies x_n \in U, x_n \in V$

But $U \cap V = \varnothing$.

From that contradiction we can see that there can be no such two distinct $l$ and $m$.

Hence the result.

$\blacksquare$