Convergent sequences might not have max or min

  • #1
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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.
 

Answers and Replies

  • #2
Yes, it is sufficient to show that inf(S) = 0 but inf(S) does not belong to S.
 
  • #3
member 587159
This works. Now, what simple thing can you do to provide an example for the other assertion?
 
  • #4
Ray Vickson
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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?)
 
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  • #5
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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##.
 
  • #6
WWGD
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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.....
 
  • #7
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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  • #8
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.
 

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