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

# Uniqueness of convergence, proposition

Let $$(a_n)_{n=m}^{\infty}$$ be a sequence of reals, and let $$L \ne L'$$ be two distinct real numbers. Then $$(a_n)_{n=m}^{\infty}$$ cannot converge to both $$L$$ and $$L'$$.

Proof
If $$L \ne L'$$, then there exists a positive real $$d = dist(L, L') = | L - L'|$$. As $$(a_n)_{n=m}^{\infty}$$ converges to $$L$$, we can choose a real $$0 < \varepsilon < \frac{d}{3}$$ and there exists an integer $$N_1 > m$$ such that $$(a_n)_{n=N_1}^{\infty}$$ is ε-close to $$L$$. Similarly, for $$L'$$ for the same $$\varepsilon$$, there. exists an integer $$N_2 > m$$ such that $$(a_n)_{n=N_2}^{\infty}$$ is ε-close to $$L'$$. 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 ε-close to both $$L$$ and $$L'$$ for all $$k \ge M$$. As we have both $$dist(a_k, L) \le \varepsilon$$ and $$dist(a_k, L') \le \varepsilon$$, this implies, by the triangle inequality, that $$dist(L, L') \le 2\varepsilon = \frac{2}{3} dist(L, L')$$, which is a false statement, as the distance is always positive. Thus, it cannot be true that $$(a_n)_{n=m}^{\infty}$$ converges to both $$L$$ and $$L'$$.

Once it is shown that sequences of reals, if they converge to a real, must converge to a unique real, we can give that unique real a name. It will be called a limit of a sequence (there is a separate card for this definition).

Thus, a limit is simply the real for which a sequence of reals converges to.

Cauchy sequences of reals → convergence of sequences of reals → uniqueness of convergence → limit, the definition → subsumption of formal limits

Tao, Analysis I
Chapter 6