I am studying KS (Kolmogorov-Sinai) entropy of order q and it can be defined as

$$ h_q = \sup_P \left(\lim_{m\to\infty}\left(\frac 1 m H_q(m,ε)\right)\right) $$

Why is it defined as supremum over all possible partitions P and not maximum?

When do people use supremum and when maximum?

  • 5
    $\begingroup$ I am not sure either of the tags fit. I do not know the term "KS entropy" so I cannot suggest a better retag. The essence of "what is $\sup$ and what is $\max$ was answered on this website before. Here is one answer, math.stackexchange.com/questions/18605/… $\endgroup$
    – Asaf Karagila
    Jul 27, 2011 at 15:55
  • $\begingroup$ Based on a quick google search I believe [dynamical-systems] fit well into this question. I (somewhat unwillingly) left [order-theory] since your second question can be generally answered within its confines. As my previous comment suggests, the second answer was indeed answered before (many times as well), I do not vote to close as the first answer may or may not been addressed to in other questions on the site. $\endgroup$
    – Asaf Karagila
    Jul 27, 2011 at 16:01
  • $\begingroup$ @Asaf, KS stands for Kolmogorov-Sinai, I saw that question earlier, but I still don't get when people use sup and when max. It's probably coz I need to spend more time to think over all the answers there. But thanks for the link anyway. Pls feel free to retag. $\endgroup$
    – oleksii
    Jul 27, 2011 at 16:03
  • 6
    $\begingroup$ "'Sup" is a hip way of asking "How are you?" $\endgroup$
    – oosterwal
    Jul 27, 2011 at 20:06
  • $\begingroup$ @oleksii: You have received several answers, all are very good in answering the question in the title. None of which answer the question "Why is it defined as supremum over all possible partitions $P$ and not maximum?". While I respect your wish to accept the answer, it would be nice to indicate that you have at least deduced the answer of the seemingly unanswered question from the accepted answer. Otherwise, this has become a duplicate of a question that has been answered here many times before. $\endgroup$
    – Asaf Karagila
    Jul 28, 2011 at 0:21

5 Answers 5


If the maximum exists, then the supremum and maximum are the same. However sometimes the maximum does not exist, and there is no maximal element. In this case it still makes sense to talk about a least upper bound.

The classic example is the set of all rationals whose square is less than or equal to $2$. That is the set $$A=\left\{ r\in\mathbb{Q}:\ r^{2}\leq2\right\}.$$

$A$ has no maximal element, however it does have a supremum and $\sup A=\sqrt{2}$.

An even simpler example is the set of all reals that are strictly less than $2$: $$B=\left\{ r\in\mathbb{R}:\ r<2\right\}.$$ This set has no maximum since for any $x\in B$ the element $\frac{x+2}{2}$ satisfies $x<\frac{x+2}{2}<2$. However it is not hard to see that $\sup B=2$.


A maximum is the largest number WITHIN a set. A sup is a number that BOUNDS a set. A sup may or may not be part of the set itself (0 is not part of the set of negative numbers, but it is a sup because it is the least upper bound). If the sup IS part of the set, it is also the max.


Consider for example the set $X = (0,1)$. Then $\sup X=1$ but $\max X$ does not exist.

Generally, for a set $X\subset {\mathbb R}$, we say $x=\max X$ if $\color{blue}{x\in X}$ and $$\forall y\in X, y\leq x.$$

We have $x=\sup X$ if $$\forall y\in X, y\leq x$$ and $$\forall\epsilon>0,\quad\exists y\in X, \quad\text{s.t.}\quad y>x-\epsilon.$$

  • $\begingroup$ By analogous reasoning $ x=\inf X $ if and only if $\forall y\in X, y \ge x$ and $\forall\epsilon>0,~\exists y\in X, ~\text{s.t.} ~y < x + \epsilon$. $\endgroup$
    – john
    Apr 20, 2022 at 23:42

Supremum need not be attained while maximum is. When maximum exist, maximum=supremum.


The set of all negative numbers has a sup, which is $0$, but not a max.

$0$ is the smallest number that no negative number can exceed.

  • $\begingroup$ Someone down-voted this. Can they explain their objection? $\endgroup$ Sep 27, 2016 at 7:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .