Convergent Real Sequence has Unique Limit/Proof 1
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Theorem
Let $\sequence {s_n}$ be a real sequence.
Then $\sequence {s_n}$ can have at most one limit.
Proof
Aiming for a contradiction, suppose that $\sequence {s_n}$ converges to $l$ and also to $m$.
That is, suppose that:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
and:
- $\ds \lim_{n \mathop \to \infty} x_n = m$
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:
\(\ds \size {l - m}\) | \(=\) | \(\ds \size {l - s_N + s_N - m}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {l - s_N} + \size {s_N - m}\) | 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 $l = m$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Proposition $1.2.3$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.22$