Convergent Sequence in Normed Vector Space has Unique Limit

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Theorem

Let $\struct {X, \norm {\,\cdot\,}}$ be a normed vector space.

Let $\sequence {x_n}$ be a sequence in $\struct {X, \norm {\,\cdot\,} }$.


Then $\sequence {x_n}$ can have at most one limit.


Proof

Aiming for a contradiction, suppose $\displaystyle \lim_{n \to \infty} x_n = L_1$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = L_2$, where $L_1 \ne L_2$.

Let $\displaystyle \epsilon = \frac {\norm {L_1 - L_2} } 3$.

From properties of norm it follows that $\epsilon > 0$.

By definition:

$\exists N_1 \in \N : \forall n > N_1 : \norm {x_n - L_1} < \epsilon$
$\exists N_2 \in \N : \forall n > N_2 : \norm {x_n - L_2} < \epsilon$

Choose $n > N_1 + N_2$.

Then:

\(\displaystyle \norm {L_1 - L_2}\) \(=\) \(\displaystyle \norm {L_1 - x_n + x_n - L_2}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {L_1 - x_n} + \norm {x_n - L_2}\) Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \epsilon + \epsilon\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 3 \norm {L_1 - L_2}\)

which implies that:

$\displaystyle 1 < \frac 2 3$

This is a contradiction.

Hence $L_1 = L_2$.

$\blacksquare$


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