# A calculus of the absurd

##### 19.1.2 A sequence can only converge to one value

Before proving this, it’s helpful to formulate precisely what we want to prove.

What 139139 Note that there exists is the negation of for every. It helps me to imagine a haystack. If you tell me that “there exists a needle in this haystack”, then the only way I can disprove you is to search through all of (for every) the haystack and show that there is no needle. If I say “for every piece of hay in this haystack, none of them is hiding a needle”, then you can disprove me by showing that at least one piece of hay is hiding a needle.we want to show is that for every converging sequence, there exists a unique value to which it converges. The only way this could not be true is if there exists at least one sequence which converges to more than one value.

To prove this by contradiction, let us assume that there is indeed a sequence $$\{x_n\}$$ which converges to both $$a$$ and $$b$$, where $$a \ne b$$. In this case we have that for all $$\epsilon > 0$$ that as $$n \to \infty$$,

\begin{align} \abs {x_n - a} < \epsilon \text { and } \abs {x_n - b} < \epsilon \end{align}

We would like to show that $$a=b$$, or that $$a-b=0$$ (as this would be a contradiction). We know that

$$\abs {a - b} = \abs {a - x_n + x_n - b}$$

Using the triangle inequality we can write that

140140 Unfortunately, this equation is a bit ambiguous - what I’m trying to say is that everything on the left-hand side is less than or equal to everything on the right-hand side and that everything on the right-hand side is equal.

\begin{align} \abs {a - x_n + x_n - b} &\leqq \abs {a - x_n} + \abs {x_n - b} \\ &= \epsilon + \epsilon \\ &= 2 \epsilon \end{align}

Therefore, overall we have that

$$\abs {a - b} \leqq 2\epsilon$$

However, we can pick any value which is greater than $$0$$ for $$\epsilon$$ here. If as we are assuming, $$a\ne b$$ then for any constant $$c>0$$ we also have that $$c \abs {a-b}$$ is greater than $$0$$. If we set $$c$$ to something less than $$\frac {1}{2}$$ (e.g. $$\frac {1}{3}$$) then we would have that

$$\abs {a - b} \leqq \frac {2}{3} \abs {a-b}$$

This is definitely not true, and hence we have derived a contradiction; thus there can only be one.