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If I have a set: $S = \{ k_1, k_2,...,k_n,... : k_i < \infty \}$, which means that $S$ include infinite many elements, but each element is a finite real number. In this case does $\max(S)$ or $\sup(S)$ exit, and is finite?

I can not find useful theorem to clarify this. Any help appreciate.

kingW3
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Amanda
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    Think about $\left{\frac 1 n \colon n\in \mathbb N\right}$. – Git Gud Jan 14 '15 at 20:01
  • By the way you describe $S$, it looks like you want it to have a countably infinite number of elements. Thus $S$ may not take the form of any interval, open or closed or otherwise, on the Real line. Is this correct? – chharvey Jan 14 '15 at 20:10
  • yes, S contain countably infinite number of elements, and all elements are finite real numbers. – Amanda Jan 14 '15 at 20:12
  • Thanks for clarifying. You should specify this in your question so that users won't give examples such as $(a,b)$ or $[a,b]$, as each of those contains an uncountably infinite number of points. – chharvey Jan 14 '15 at 20:14
  • Should it be s/exit/exist? – SOFe Sep 18 '19 at 08:21

2 Answers2

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Some examples:

$$ \mathbb{N} = \left\{1, 2, 3, 4, 5, \dots\right\} $$

$\mathbb{N}$ does not have a maximum, nor does it have a supremum.

$$ T = \left\{1-\frac{1}{n} : n \in \mathbb{N} \right\}$$

$T$ does not have a maximum but does have a supremum: $\sup(T) = 1$.

$$ U = \left\{\frac{1}{n} : n \in \mathbb{N} \right\}$$

$U$ has a maximum and a supremum, which are necessarily the same (a maximum of a set must be its least upper bound): $\max(U) = \sup(U) = 1$.

chharvey
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  • Thanks for your answer, under your examples, such a set may or may not have a supremum. While I am not sure if the condition "every element is finite real number" is considered. In your first example S= { 1,2,3,4,5,...}, can we say all element are finite? – Amanda Jan 14 '15 at 20:17
  • Yes. By the pattern, $1, 2, 3, 4, 5, \dots$, each is a natural number. Natural numbers are known to be "finite," that is, informally, "less than infinity." – chharvey Jan 14 '15 at 20:19
  • @Amanda for another example, take the set of all prime integers. There are countably infinitely many, and each prime integer is "finite", yet there is no maximum or supremum of the set. – chharvey Jan 14 '15 at 20:25
  • Thanks! So the answer is negative. It is not a happy conclusion for me, as I hope it is positive so I can use this in my proof. The good thing is that I got it does not work, I should try something else. Thanks a lot everyone! – Amanda Jan 14 '15 at 20:28
  • Glad you solved the problem. Don't forget to upvote and accept the answer you think is the most helpful. – chharvey Jan 14 '15 at 20:29
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An infinite set which is bounded above always has a supremum by the completeness of $\mathbb{R}$; i.e. any set which has an upper bound has a least upper bound. A set $A \subseteq \mathbb{R}$ is said to be bounded above if there is $M \in \mathbb{R}$ such that $a \leq M$ for all $M$.

Examples:

  1. The infinite set $\mathbb{N}$ is not bounded above so it does not have a supremum.
  2. The set $[0, 1]\cap\mathbb{Q}$ is bounded above; for example, taking $M = 1$ we see that $a \leq M$ for every $a \in [0, 1]\cap\mathbb{Q}$. Therefore, $[0, 1]$ has a supremum.
  3. The set $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ is bounded above; for example, taking $M = 0$ we see that $a \leq 0$ for every $a \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$. Therefore, $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ has a supremum.

An element $x \in A \subseteq \mathbb{R}$ is said to be maximal (or just a maximum of $A$) if $a \leq x$ for all $a \in A$. The difference between this definition and the related definition above is that $x$ belongs to $A$ whereas $M$ need not belong to the set in question. A finite subset always has a maximum, but an infinite set may or may not, even if it is bounded above.

Examples:

  1. The set $[0, 1]\cap\mathbb{Q}$ has a maximum, namely $1$ because for any $a \in [0, 1]\cap\mathbb{Q}$, $a \leq 1$.
  2. The set $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ does not have a maximum, because for any $x \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$, there is $a \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$ such that $x < a$. More precisely, if $x = -\frac{1}{n}$, then $a = -\frac{1}{2n}$ satisfies $x < a$.
Michael Albanese
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  • Because the OP describes $S$ as $\left{k_1, k_2, \dots, k_n\right}$, it seems that $S$ must contain a countably infinite number of points, thus intervals such as $[a,b]$ or $(a,b)$ are not good examples. – chharvey Jan 14 '15 at 20:12
  • Well, the issues involved here really have nothing to do with countable vs. uncountable, but I take your point. – Michael Albanese Jan 14 '15 at 20:14
  • Agreed, but I think from anyone asking this type of question, whether the issue might have some involvement is unclear. – chharvey Jan 14 '15 at 20:15