Unable to come up with correct bounds for showing sequence convergence. Analysis I, Terence Tao, Theorem 6.1.19, part (g)

Question

I was trying to prove the following (part (g) of the Theorem):

Theorem 6.1.19 (Limit Laws). Let $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ be convergent sequences of real numbers, and let $x,y$ be the real numbers $x := \lim_{n \to \infty}a_n$ and $y := \lim_{n \to \infty}b_n$.

(g) The sequence $(\text{max}(a_n,b_n))_{n=m}^\infty$ converges to $\text{max}(x,y)$; in other words,

$$ \lim_{n \to \infty} \text{max}(a_n,b_n)=\text{max}(\lim_{n \to \infty}a_n,\lim_{n \to \infty}b_n) $$

I have written out my attempt below, but the proof is incomplete. I have laid out two questions within the writing below that I'm hoping someone could help me with. I am aware there are solution to this already online, but I wanted to see if my attempt is worth fixing if possible.

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Proof attempt.

By the hypothesis of the theorem, we have that for every $\varepsilon > 0$, there exist positive integers $N_1,N_2 \geq m$ such that

$$ |a_n-x| \leq \varepsilon, \ \text{for all} \ n \geq N_1 $$

and

$$ |b_n-y| \leq \varepsilon, \ \text{for all} \ n \geq N_2 $$

Assume, without loss of generality, that $x \leq y$, so that $\text{max}(x,y)=y.$

We want to show that for every $\varepsilon > 0$, there exists a positive integer $N \geq m$ such that $|\text{max}(a_n,b_n) - y| \leq \varepsilon$, for all $n \geq N$. Well, we have that

\begin{align*} |\text{max}(a_n,b_n) - y| &= |\text{max}(a_n,b_n)-b_n+b_n-y| \\ &\leq |\text{max}(a_n,b_n) - b_n| + |b_n - y| \\ \end{align*}

I want to bound $|\text{max}(a_n,b_n) - b_n|$. Well, for $n\geq m$, we have two cases depending on $n$. We could have $\text{max}(a_n,b_n)=b_n$ or $\text{max}(a_n,b_n)=a_n$. In the former case, we'd have $|\text{max}(a_n,b_n) - b_n|=|b_n-b_n|=0$. In the latter case, we'd have $|\text{max}(a_n,b_n) - b_n|=|a_n-b_n|\geq0$. Thus, regardless which case it is, we'll always have $|\text{max}(a_n,b_n) - b_n| \leq|a_n-b_n|$.

Question 1. Is my reasoning in the above paragraph valid?

Continuing where we left off, we then have

\begin{align*} |\text{max}(a_n,b_n) - y| &= |\text{max}(a_n,b_n)-b_n+b_n-y| \\ &\leq |\text{max}(a_n,b_n) - b_n| + |b_n - y| \\ &\leq |a_n-b_n| + |b_n-y| \\ &= |a_n-x + x-b_n| + |b_n-y| \\ &\leq |a_n-x| + |x-b_n| + |b_n-y| \\ &= |a_n-x| + |b_n-y| + |b_n-y+y-x| \\ &\leq |a_n-x| + |b_n-y| + |b_n-y| + |y-x| \\ &\leq |a_n-x| + |b_n-y| + |b_n-y| + |y-x| + 1 \end{align*}

With $\varepsilon > 0$ and $|y-x| + 1 > 0$, we can use the Archimedean Property to find a positive integer $M$ such that $|y-x| + 1 < M \varepsilon$. This implies that

\begin{align*} |\text{max}(a_n,b_n) - y| & \leq |a_n-x| + |b_n-y| + |b_n-y| + M \varepsilon \end{align*}

Question 2. At this point, I got stuck. I have been trying to find bounds for $|a_n-x|$ and $|b_n-y|$ such that we still have $\varepsilon > 0$ and that the right side of the inequality is either eqaul to $\varepsilon$ or less than it. But no matter what I thought of, nothing worked. Is this a lost cause? Should I think of a different strategy?

Answer 1

Your idea appears to be to use $|\max(a_n,b_n)-y| \leq |a_n-b_n|+|b_n-y|$ to show $\max(a_n,b_n)\to y.$ The derivations of this inequality is correct. However, the method will not work in general, for example if $a_n = 1$ and $b_n = 2$ for all $n,$ then the term $|a_n-b_n|=1$ cannot be made smaller than any $\epsilon<1.$.