Convergent Sequence in Hausdorff Space has Unique Limit

Theorem

Let $T = \struct {S, \tau}$ be a Hausdorff space.

Let $\sequence {x_n}$ be a convergent sequence in $T$.

Then $\sequence {x_n}$ has exactly one limit.

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

Aiming for a contradiction, suppose $l$ and $m$ are both limits of $\sequence{x_n}$.

As $T$ is Hausdorff, there exist $U, V \in \tau$ such that:

$l \in U, m \in V$ and $U \cap V = \O$

Then, from the definition of convergent sequence:

$\ds \exists N_U \in \R: \, $

$\ds n > N_U$

$\implies$

$\ds x_n \in U$

$\ds \exists N_V \in \R: \, $

$\ds n > N_V$

$\implies$

$\ds x_n \in V$

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

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

But $U \cap V = \O$.

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

Hence the result.

$\blacksquare$

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.2$