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.


8 Answers 8


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 $\max (a_n,b_n)\to a = \max(a,b)$.

Separate your task into two cases:

  1. If $a=\lim_{n\to\infty} a_n \neq \lim_{n\to\infty} b_n=b$, then you can assume, without loss of generality, that $a>b$. In this case, you can show that from some $n$ on, $a_n > b_n$ and thus that $\lim_{n\to\infty}\max(a_n,b_n) = \lim_{n\to\infty} a_n = a$
  2. If $a=b$, then it is best to go to the definition of the limit using epsilons to prove that $a$ must be the limit of the max sequence.

Note that $$\max(A,B) = \max(B-A,0)+A$$

for all $A,B$.

This reduces the problem to showing that for a convergent sequence $z_n$ with $\lim\limits_{n\to \infty }z_n=z$ we have

$$ \lim_{n\to \infty } \max(z_n,0) = \max(z,0).$$


$$ |\max(C,0) - \max(D,0)| \leq |C-D| $$

for all $C,D$, we have

$$ |\max(z_n,0) - \max(z,0)| \leq |z_n-z| $$

for all $n$, and using squeeze theorem the result follows.

Note: This is the same strategy as the one used to show

$$ \lim_{n\to \infty } |z_n| = |z| $$ where we exploit the reverse triangle inequality:

$$ ||z_n| - |z|| \leq |z_n-z| $$

for all $n$.


Suppose $x\neq y$, WLOG assume $x> y$, let $\epsilon= \frac{x-y}{2}$, by definition of convergence, you can show eventually $a_n>x-\epsilon, b_n<y+\epsilon$, hence $\max(a_n,b_n)=a_n$ eventually, hence $$\lim_{n\to\infty}\max(a_n,b_n)=\lim_{n\to\infty}a_n=x=\max(x,y)$$

Suppose $x=y$, then $\forall \epsilon>0$, by definition of convergence, you can show eventually $|a_n-x|<\epsilon, |b_n-a|<\epsilon$, hence $|\max(a_n,b_n)-x|<\epsilon$ eventually

  • $\begingroup$ This makes two more assumtions that are implicit: the limits of both sequences are not the same and the limit of $a_n$ is greater than the limit of $b_n$. $\endgroup$
    – 5xum
    Commented Oct 28, 2014 at 12:54
  • $\begingroup$ @5xum Thanks. I thought the case for $x=y$ is easy. I will add more details for completeness. $\endgroup$
    – John
    Commented Oct 28, 2014 at 12:59

Let $\epsilon>0$. Then there exists an $N$ such that $x-\epsilon\leq a_n\leq x+\epsilon$ and $y-\epsilon\leq b_n\leq y+\epsilon$ for each $n\geq N$. Thus $$ \max(x,y)-\epsilon=\max(x-\epsilon,y-\epsilon)\leq\max(a_n,b_n)\leq\max(x+\epsilon,y+\epsilon)=\max(x,y)+\epsilon $$ for each $n\geq N$.

Extra. The proof for $\min$ follows from the above by using $\min(x,y)=-\max(-x,-y)$ for $x,y\in\mathbf R$.


Following notation and terminology as in Terence Tao's "Analysis I", as it seems this question is taken from Exercise 6.1.8 (p. 132) where the reader is asked to prove Theorem 6.1.19 (p. 130).

We need to show that for any $\epsilon > 0$, the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^\infty$ is eventually $\epsilon$-close to $\max(x, y)$. Let $\epsilon > 0$ be arbitrary.

We know that we can find two natural numbers $N_1, N_2\geq m$ such that $$ \begin{align*} |a_n - x| \leq \epsilon \quad \text{and} \quad |b_n - y| \leq \epsilon \end{align*} $$ for all $n \geq \max(N_1, N_2)$. Let $N = \max(N_1, N_2)$.

We have two cases to consider: a) when $x = y$ and b) when $x \neq y$.

Assume first that $x = y$. In that case, $\max(x, y) = x = y$, so our task reduces to showing that the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $x$.

We know that for all $n \geq N$ we have both $|a_n - x| \leq \epsilon$ and $|b_n - x| \leq \epsilon$ (note the use of $x$ instead of $y$ under the assumption $x = y$). Hence, no matter whether $\max(a_n, b_n)$ is equal to $a_n$ or $b_n$ we are guaranteed that $|\max(a_n, b_n) - x| \leq \epsilon$ for all $n \geq N$. So the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $\max(x, y)$.

Now, for the second case assume that $x \neq y$. By the trichotomy of order on the real numbers we must have either $x < y$ or $x > y$. We really need to check both cases, but it turns out that the arguments used for one case can be tailored to fit with the other case. So we assume that $x > y$.

This assumption tells us that $\max(x, y) = x$. We therefore need to show that the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $x$. Again, we know that for all $n \geq N$ $$ \begin{align*} |a_n - x| \leq \epsilon \quad \text{and} \quad |b_n - y| \leq \epsilon. \end{align*} $$

More specifically, we can set $\epsilon = \tfrac{x - y}{2}$ (note that since $x > y$ by assumption, this is a positive number, so $\epsilon > 0$.) Using the fact (shown in Exercise 5.4.6) that $$ \begin{align*} |a_n - x| \leq \epsilon \implies x - \epsilon \leq a_n \leq x + \epsilon \\ |b_n - y| \leq \epsilon \implies y - \epsilon \leq b_n \leq y + \epsilon \end{align*} $$ we want to show that, at some point $a_n \geq b_n$, because then at some point $\max(a_n, b_n) = a_n$. Using our epsilon from above, we have that $b_n \leq y + (x - y)/2 = (x + y)/2$. Similarly, we have that $x - (x-y)/2 = (x + y)/2 \leq a_n$. So, $$ b_n \leq \frac{x+y}{2} \leq a_n $$ for all $n \geq N$. Hence for all $n \geq N$ we have $|\max(a_n, b_n) - x| = |a_n - x| \leq \epsilon$. Consequently, the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $\max(x, y)$.

This concludes the proof.


Note that for any $n$, $$\max(a_n,b_n)=\frac{a_n+b_n+\vert a_n-b_n\vert}{2}.$$So $$\lim \max(a_n,b_n)=\frac{\lim a_n+\lim b_n+\vert \lim a_n-\lim b_n\vert}{2}=\max(\lim a_n,\lim b_n)$$

  • 1
    $\begingroup$ That’s a formula for the min not the max. I suggested an edit but, according to two reviewers, it does not improve the quality of the post. $\endgroup$
    – Taylor
    Commented Jul 16, 2021 at 23:14

The following result is from John M. Lee's book about Introduction to Topological Manifolds.

Gluing Lemma: Let $X$ and $Y$ be topological spaces, and let $\left\{A_i\right\}$ be either an arbitrary open cover of $X$ or a finite closed cover of $X$. Suppose that we are given continuous maps $f_i\colon A_i\to Y$ that agree on overlaps: $f_i|_{A_i\cap A_j}=f_j|_{A_i\cap A_j}$. Then there exists a unique continuous map $f\colon X\to Y$ whose restriction to each $A_i$ is equal to $f_i$.

In our problem: for topological spaces take $X=\mathbb{R}^2$, $Y=\mathbb{R}$; for closed cover, take $A_1=\left\{(x,y)\in\mathbb{R}^2:\:x\le y\right\}$, $A_2=\left\{(x,y)\in\mathbb{R}^2:\:y\le x\right\}$; and for maps take $f_1(x,y)=y$ for $(x,y)\in A_1$ and $f_2(x,y)=x$ for $(x,y)\in A_2$.

Notice that for every $(x,y)\in \mathbb{R}^2$,

\begin{align*} \max\left\{x,y\right\}&=\begin{cases}y&\text{if }x\le y\\x&\text{if }y\le x\end{cases}=\begin{cases}f_1(x,y)&\text{if }(x,y)\in A_1\\f_2(x,y)&\text{if }(x,y)\in A_2\end{cases}. \end{align*}

Therefore $f(x,y)\equiv\max\left\{x,y\right\}$, by uniqueness, and so it is continuous.

So, whenever $x_n\to x$ and $y_n\to y$, we have $$\max\left\{x_n,y_n\right\}=f(x_n,y_n)\to f(x,y)=\max\left\{x,y\right\},$$ as $n\to \infty$.


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