Exercise 2.2.6

Question

extbf{Theorem 2.2.7 (Uniqueness of Limits).} 	extit{The limit of a sequence, when it exists, must be unique.}

Prove Theorem 2.2.7. To get started, assume $\left(a_{n}ight) ightarrow a$ and also that $\left(a_{n}ight) ightarrow b$. Now argue $a=b$

Ulisse Mini · Accepted Answer

If $a 
e b$ then we can set $\epsilon$ small enough that having both $|a_n - a| < \epsilon$ and $|a_n - b| < \epsilon$ is impossible. Therefore $a = b$.

(Making this rigorous is trivial and left as an exercise to the reader)

Richard Ganaye · Answer

As a faithful reader, I write this proof.

\begin{proof} 
Assume for contradiction that $a 
e b$. Take $\varepsilon = \frac{|b-a|}{2024}$ (*). Since $\lim\limits_{n 	o \infty} a_n =a$ and $\lim\limits_{n 	o \infty} a_n =b$, there are integers $N_1,N_2  \in \mathbf{N}$ such that
\begin{align*}
\forall n \in \mathbf{N},\ n\geq N_1 \Rightarrow |a_n -a |< \varepsilon,\
\forall n \in \mathbf{N},\ n\geq N_2 \Rightarrow |a_n -b|< \varepsilon.
\end{align*}
Define $N = \max(N_1,N_2)$. For all $n \in \mathbf{N}$, if $n\geq N$, then $n\geq N_1$ and $n\geq N_2$, so
$$|a_n -a |< \varepsilon,\qquad |a_n - b | < \varepsilon.$$
In particular, for $n = N = \max(N_1,N_2)$, we obtain
$$|b-a| \leq |b - a_n| + |a_n - a| <2  \varepsilon = 2\frac{|b-a|}{2024}.$$
Because $|b-a| 
e 0$, $1 < \frac{2}{2024}$ so $2024 < 2$: this is a contradiction. This proves $a = b$ and the unicity of the limit.
\end{proof}
(*) This little trick allows the teacher to renew his course every year. Personally, this caused me trouble only during the year 0001.

Exercise 2.2.6

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Exercise 2.2.6

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