Convergent Sequence in Hausdorff Space has Unique Limit

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ 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, $\exists U \in \tau: l \in U$ and $\exists V \in \tau: m \in V$ such that $U \cap V = \varnothing$.


Then, from the definition of convergent sequence:

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


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

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

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.

$\blacksquare$


Sources