Real Sequence has One Limit at Most

Theorem

Let $\left \langle {s_n} \right \rangle$ be a real sequence.

Then $\left \langle {s_n} \right \rangle$ can have at most one limit.

Proof

Suppose that $\left \langle {s_n} \right \rangle$ converges to $l$ and also to $m$.

That is, suppose $\displaystyle \lim_{n \to \infty} x_n = l$ and $\displaystyle \lim_{n \to \infty} x_n = m$.

Assume that $l \ne m$, and let:

$\epsilon = \dfrac {\left\vert{l - m}\right\vert} 2$

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

Therefore, since $\left \langle {s_n} \right \rangle \to l$:

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

Similarly, since $\left \langle {s_n} \right \rangle \to m$:

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

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

We have:

	\(\displaystyle \)	\(\displaystyle \left\vert{l - m}\right\vert\)	\(=\)	\(\displaystyle \left\vert{l - s_N + s_N - m}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\vert{l - s_N}\right\vert + \left\vert{s_N - m}\right\vert\)	\(\displaystyle \)	by the Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle 2 \epsilon\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{l - m}\right\vert\)	\(\displaystyle \)

This constitutes a contradiction.

Therefore, it must be that $l = m$.

$\blacksquare$

Alternative Proof

From the fact that the real number line is a metric space, we can directly use Sequence in Metric Space has One Limit at Most‎.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Proposition $1.2.3$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.22$