Convergent sequences might not have max or min

[Mr Davis 97]

Homework Statement

Prove or produce a counterexample: If $\{a_n\}_{n=1}^\infty\subset \mathbb{R}$ is convergent, then
$\min (\{a_n:n\in\mathbb{N}\} )$ and
$\max (\{a_n:n\in\mathbb{N}\} )$
both exist.

Homework Equations

The Attempt at a Solution

I will produce a counterexample.
Let $a_n = \frac{1}{n}$. First we will show that $a_n$ is converges to 0.
Let $\epsilon > 0$. By the Archimedian principle there exists an $N \in \mathbb{N}$ such that $\frac{1}{N} < \epsilon$. If $n \ge N$, then $|a_n - 0| = |\frac{1}{n} - 0| = \frac{1}{n} \le \frac{1}{N} < \epsilon$. So $\lim a_n = 0$.

Let $S = \{\frac{1}{n}:n\in\mathbb{N}\}$. Then $\max(S) = \sup(S) = 1$. However, the set does not have a minimum, and this is where I'm stuck. Do I need to show that $\inf(S) = 0$ and that $0 \not \in S$? Clearly there is no minimum, but I am not sure how to show this rigorously.

 

[ConfusedMonkey]

Yes, it is sufficient to show that inf(S) = 0 but inf(S) does not belong to S.

 

[member 587159]

This works. Now, what simple thing can you do to provide an example for the other assertion?

 

[Ray Vickson]

This works. Now, what simple thing can you do to provide an example for the other assertion?

Since the question asked whether both a max and a min exist, he has disproved that by showing there need not be a min. A similar example could be given about the max.

However, a more interesting (and seemingly much harder) case is whether one can find a single example in which both a max and a min fail to exist. (In other words, must every convergent sequence have either a max or a min?)

 

[Mr Davis 97]

This works. Now, what simple thing can you do to provide an example for the other assertion?

I'm actually not sure how to show that $\inf(S) = 0$. Does it go something like this?

1) 0 is a lower bound. Since clearly $\forall n \in \mathbb{N}$, $0 \le \frac{1}{n}$.
2) Let $b$ be a lower bound for $S$. Then $\forall n \in \mathbb{N}$, $b \le \frac{1}{n}$... This is where I'm stuck. I don't see how to conclude $b \le 0$.

 

[WWGD]

I'm actually not sure how to show that $\inf(S) = 0$. Does it go something like this?

1) 0 is a lower bound. Since clearly $\forall n \in \mathbb{N}$, $0 \le \frac{1}{n}$.
2) Let $b$ be a lower bound for $S$. Then $\forall n \in \mathbb{N}$, $b \le \frac{1}{n}$... This is where I'm stuck. I don't see how to conclude $b \le 0$.

One way of seeing is that since 0 is the limit, points must ultimately become indefinitely small. If {1/n} had a minimum >0 then...

 

[member 587159]

I'm actually not sure how to show that $\inf(S) = 0$. Does it go something like this?

1) 0 is a lower bound. Since clearly $\forall n \in \mathbb{N}$, $0 \le \frac{1}{n}$.
2) Let $b$ be a lower bound for $S$. Then $\forall n \in \mathbb{N}$, $b \le \frac{1}{n}$... This is where I'm stuck. I don't see how to conclude $b \le 0$.

There is more than one way:

(1) $(1/n)_n$ is a monotone decreasing bounded sequencce. Hence, it converges to its infinum by the monotone convergence theorem. But $\lim_n 1/n = 0$. It follows that the infinum must be 0.

(2) Clearly, $0 \leq 1/n$ for all n. Hence, $0$ is an upper bound of the set. Use the archimedian property to show that $0$ is also the least upper bound.

 

[member 587159]

Since the question asked whether both a max and a min exist, he has disproved that by showing there need not be a min. A similar example could be given about the max.

However, a more interesting (and seemingly much harder) case is whether one can find a single example in which both a max and a min fail to exist. (In other words, must every convergent sequence have either a max or a min?)

My intuition tells me that your last statement is true.

Let me try to give a proof. Since our sequence $(a_n)$ is convergent, it is bounded. For the sake of reaching a contradiction, assume the sequence has neither a max nor a min. Put $$R:= \{a_n\mid n\}$$

Define $m:= \inf R, M:= \sup R$, which exist by boundedness of the sequence.

We can find subsequences converging to both $m$ and $M$ (this needs justification using that $m,M \notin R$)

Because our sequence is convergent, it follows that $m=M$. This means that R is a singelton, which means R contains its maximum and minimum.

Contradiction.