Convergent Sequence in Hausdorff Space has Unique Limit

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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.


Proof

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


Aiming for a contradiction, suppose $\ds \lim_{n \mathop \to \infty} x_n = l$ and $\ds \lim_{n \mathop \to \infty} x_n = m$ such that $l \ne m$.

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.

$\blacksquare$


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