Convergent Sequence in Normed Vector Space has Unique Limit

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 $\ds \lim_{n \mathop \to \infty} x_n = L_1$ and $\ds \lim_{n \mathop \to \infty} x_n = L_2$ such that $L_1 \ne L_2$.

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

From the norm axioms 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:

\(\ds \norm {L_1 - L_2}\)

\(=\)

\(\ds \norm {L_1 - x_n + x_n - L_2}\)

\(\ds \)

\(\le\)

\(\ds \norm {L_1 - x_n} + \norm {x_n - L_2}\)

Triangle Inequality

\(\ds \)

\(<\)

\(\ds \epsilon + \epsilon\)

\(\ds \)

\(=\)

\(\ds \frac 2 3 \norm {L_1 - L_2}\)

which implies that:

$1 < \dfrac 2 3$

This is a contradiction.

Hence $L_1 = L_2$.

$\blacksquare$

Also see

• Convergent Sequence has Unique Limit

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces