Real Analysis – Limit Laws for Maximum of Sequences

Question

Let $(a_n)^{\infty}_{n=m}$ and $(b_n)^{\infty}_{n=m}$ be convergent sequences of real numbers.

Let $x$ and $y$ be the real numbers $x:=\lim\limits_{n\to\infty}a_n$ and $y:=\lim\limits_{n\to\infty}b_n$.

Show that the sequence $(\max(a_n,b_n))$ converges to $\max(x,y)$; in other words: $$\lim_{n\to\infty}\max(a_n,b_n)=\max\bigl(\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n\bigr)$$

I was not able to prove it and would appreciate your help.

Answer 1

Assume that $a_n \to a$ and $b_n \to b$.

The most simple is to split this in two cases.

1. if $a\neq b$: Assume without loss of generality that $a<b$. Let $\epsilon = \frac {b-a}2$. You can find $N_{1,2}$ such as \begin{align} n > N_1 &\implies |a_n - a|<\epsilon\\ n > N_2 &\implies |b_n - b|<\epsilon\\ \implies [n\ge \max(N_1, N_2) &\implies |a_n - a|<\epsilon, |b_n - b|<\epsilon] \end{align} Now if $n\ge \max(N_1, N_2)$: $$a_n<a+\epsilon=b-\epsilon<b_n\\ $$ so $\max (a_n, b_n) = b_n \to b = \max (a,b)$.
2. If $a=b$: let $\epsilon>0$. You can find $N_{1,2}$ such as \begin{align} n > N_1 &\implies |a_n - a|<\epsilon\\ n > N_2 &\implies |b_n - a|<\epsilon\\ \implies [n\ge \max(N_1, N_2) &\implies |a_n - a|<\epsilon, |b_n - a|<\epsilon]\\ \implies [n\ge \max(N_1, N_2) &\implies |\max (a_n,b_n) - a|<\epsilon]\\ \end{align} so $\max (a_n,b_n)\to a = \max(a,b)$.