Math and science::Analysis::Tao::06. Limits of sequences

Proof

If $L\ne L^{\prime }$, then there exists a positive real $d=dist(L,L^{\prime })=|L-L^{\prime }|$. As $(a_n{)}_{n=m}^{\mathrm{\infty }}$ converges to $L$, we can choose a real $0<\epsilon <\frac{d}{3}$ and there exists an integer $N_1>m$ such that $(a_n{)}_{n=N_1}^{\mathrm{\infty }}$ is [...] $L$. Similarly, for $L^{\prime }$ for the same $\epsilon$, there. exists an integer $N_2>m$ such that $(a_n{)}_{n=N_2}^{\mathrm{\infty }}$ is [...] $L^{\prime }$. Thus, there exists an $M=max(N_1,N_2)$ (or just $M=N_1+N_2$, if you don't want to use $max$), such that $a_k$ is [...] for all $k\ge M$. As we have both $dist(a_k,L)\le \epsilon$ and $dist(a_k,L^{\prime })\le \epsilon$, this implies, by the [...], that $dist(L,L^{\prime })\le 2\epsilon =\frac{2}{3}dist(L,L^{\prime })$, which is a false statement, as the distance is always positive. Thus, it cannot be true that $(a_n{)}_{n=m}^{\mathrm{\infty }}$ converges to both $L$ and $L^{\prime }$.