Convergent Real Sequence has Unique Limit

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Theorem

Let $\sequence {s_n}$ be a real sequence.

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

Proof 1

Aiming for a contradiction, suppose that $\sequence {s_n}$ converges to $l$ and also to $m$.

That is, suppose that:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$

and:

$\displaystyle \lim_{n \mathop \to \infty} x_n = m$

Without loss of generality, assume that $l \ne m$.

Let:

$\epsilon = \dfrac {\size {l - m} } 2$

As $l \ne m$, it follows that $\epsilon > 0$.

As $\sequence {s_n} \to l$:

$\exists N_1 \in \N: \forall n \in \N: n > N_1: \size {s_n - l} < \epsilon$

Similarly, since $\sequence {s_n} \to m$:

$\exists N_2 \in \N: \forall n \in \N: n > N_2: \size {s_n - m} < \epsilon$

Now set $N = \max \set {N_1, N_2}$.

We have:

 $\displaystyle \size {l - m}$ $=$ $\displaystyle \size {l - s_N + s_N - m}$ $\displaystyle$ $\le$ $\displaystyle \size {l - s_N} + \size {s_N - m}$ Triangle Inequality for Real Numbers $\displaystyle$ $<$ $\displaystyle 2 \epsilon$ $\displaystyle$ $=$ $\displaystyle \size {l - m}$

This constitutes a contradiction.

It follows from Proof by Contradiction that $l = m$.

$\blacksquare$

Proof 2

We have that the real number line is a metric space

The result then follows from Convergent Sequence in Metric Space has Unique Limit‎.

$\blacksquare$