Limits are Unique

Unique Limits

If a sequence has a limit, then the limit is unique.

Proof
Let $a_{n} \to a$ and $a_{n} \to b$ are both true. We want to show that $a = b$ . We know that then $\forall ε > 0 \exists N \in N (| a_{n} - a | < ε)$ and $ε > 0 \exists N \in N (| a_{n} - b | < ε)$ .

Here let $ε > 0$ (I know we aren't proving convergence, but we want to prove that $| a - b | < ε$ since then $a = b$ . If you're confused see Least Upper Bounds and Greatest Upper Bounds and namely the Suprema Lemma for context). Since $ε > 0$ , and thus $\frac{ε}{2} > 0$ then there are $N_{1} \in N$ such that for all $n \geq N_{1}$ that $| a_{n} - a | < \frac{ε}{2}$ (why not just $ε$ ? Later we'll have to add two $ε$ 's in the worst case, so then we need half of this to show that it's still possible). Similarly there is a $N_{2} \in N$ where for all $n \geq N_{2}$ then $| a_{n} - b | < \frac{ε}{2}$ .

Now with that then:

\begin{aligned} | a - b | & = | (a - a_{n}) + (a_{n} - b) | \\ \leq | a - a_{n} | + | a_{n} - b | & Triangle Inequality \\ = | a_{n} - a | + | a_{n} - b | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

Thus $| a - b | < ε$ so then $a = b$ as required.
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