Convergent Real Sequence has Unique Limit/Proof 1

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

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


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

That is, suppose that:

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


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

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


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

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:

\(\ds \size {l - m}\) \(=\) \(\ds \size {l - s_N + s_N - m}\)
\(\ds \) \(\le\) \(\ds \size {l - s_N} + \size {s_N - m}\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds 2 \epsilon\)
\(\ds \) \(=\) \(\ds \size {l - m}\)

This constitutes a contradiction.

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