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Let $s_n$ be a real sequence. Then $s_n$ can have at most one limit.

The proof is listed here. (Link to the current revision.)

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

Similarly, since $\left \langle {s_n} \right \rangle \to m$: $$\exists N_2 \in \mathbb N: \forall n \in \mathbb N: n > N_2: \left\vert{s_n - m}\right\vert < \epsilon$$

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

We have:

\begin{align*} \left\vert{l - m}\right\vert &=\left\vert{l - s_N + s_N - m}\right\vert\\ &\le\left\vert{l - s_N}\right\vert + \left\vert{s_N - m}\right\vert \qquad\text{by the Triangle Inequality}\\ &<2 \epsilon\\ &=\left\vert{l - m}\right\vert \end{align*}

This constitutes a contradiction.

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

My question is when they set $\epsilon = \frac{|l-m|}{2}$, and thereby show that $|l-m| < |l-m|$, how does this contradict the fact that $l \neq m$? Doesn't it just show that $\epsilon$ can't be $\frac{|l-m|}{2}$? I'm confused on how the proof by contradiction works here.

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  • $\begingroup$ Because the proof began with "Assume $l \neq m$." It is that assumption that leads to a contradiction. $\endgroup$ Aug 21, 2016 at 6:44
  • $\begingroup$ But what about the assumption that $\epsilon = \frac{|l-m|}{2}$? $\endgroup$
    – Dak Song
    Aug 21, 2016 at 6:48
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    $\begingroup$ that's not assumption. It's saying, assume $l\neq m$. Then pick $\epsilon=\frac{|l-m|}{2}$, which is positive by assumption. $\endgroup$ Aug 21, 2016 at 6:49
  • $\begingroup$ Perhaps it's helpful to step back. The idea is that if the sequence gets arbitrarily close to $l$ and arbitrarily close to $m$ then $l$ and $m$ must be arbitrarily close to each other. The epsilonics is just a way of formalizing that idea. $\endgroup$
    – user4894
    Aug 21, 2016 at 6:56
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    $\begingroup$ I did not see that you wrote "how does this contradict the fact that $l\neq m$." That is another mistake you are making. Your goal is not to contradict your assumption, at least not necessarily. To prove $P$, you assume $\lnot P$ and derive any contradiction. Then $P$ follows. $\endgroup$ Aug 21, 2016 at 7:16

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This seems to be a very common misconception, namely that in a proof by contradiction you need to have a very specific contradiction. You don't. If you're able to derive $|\ell-m|< |\ell-m|$ for real $\ell,m$, that's a contradiction no matter what, there are no real numbers with that property, so you could say it contradicts your assumption that $\ell,m$ are real.

Think of the proof you linked as stating the following:

Let $(x_n) \rightarrow \ell$ and $(x_n) \rightarrow m$. If $\ell \neq m$, then there are two real numbers, namely $\ell$ and $m$ such that $|\ell-m|<|\ell-m|$.

Now, we know this conclusion is false, no such real numbers exist. So this contradicts a well known property of real numbers (really of it's a property of ordering).

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