# Convergent sequences might not have max or min

## 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.

## 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.

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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
Homework Helper
Dearly Missed
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
Gold Member
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.

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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.