Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am following Mathematical Analysis by Tom Apostol. I have a confusion here.

3. The half-open interval S = [0,1) is bounded above by 1 but has no maximum element. It's minimum element is 0.

For these sets like in Example 3, which are bounded above but have no maximum element, there is a concept which takes place of the maximum element. It is called least upper bound or supremum as is defined as follows

And there is a property called Approximation property

Theorem (Approximation property): Let S be a set of nonempty set of Real numbers with a supremum, say b = sup(S). Then for every a $$a < x \leq b$$

My question is shouldn't it be like this $ a \leq x < b$ since the set S does not contain b? Or does it imply that we approximate b to be in S?

Also if the the maximum element of the set exist, then would it be the supremum?

And finally, there is a statement

... The completeness axiom allows us to introduce irrational numbers in real number system, and it gives the real number system a property of continuity that is fundamental to many theorems in analysis.

How does completeness axiom allows us to introduce irrational number in real number system?

share|cite|improve this question
up vote 3 down vote accepted

The supremum $\sup A$ is the least upper bound, which exists for every set $A$that is nonempty and bounded from above (and usually the definition is extended so that $\sup A=+\infty$ if $A$ is not bounded from above and $\sup A=-\infty$ if $A=\emptyset$). If the set $A$ does have a maximal element, then this element is at the same time an upper bound and is obviously the least such bound, hence $\max A=\sup A$ if $\max A$ exists. For the same reason, you cannot say that definitely the least upper bound $\sup A$ is not an element of $A$. In that sense the text is misleading when it says that the supremum is a concept that is introduced only for sets that don't have a maximum. It's not a replacement, rather an extension.

The completeness is the "only" property that makes $\mathbb R$ differ from $\mathbb Q$. In $\mathbb Q$, the set $\{x\in\mathbb Q\mid x^2<2\}$ is nonempty and bounded from above (e.g. by $42$), but there is no least upper bound in $\mathbb Q$. On the other hand, completeness of $\mathbb R$ gives us he existence of a number $w:=\sup\{x\in\mathbb Q\mid x^2<2\}$ that is a least upper bound. One can easily show that $w^2=2$ (because neither $w^2<2$ nor $w^2>2$ are possible). Since there is no rational solution of $X^2-2=0$, we conclude that $w$ is irrational.

share|cite|improve this answer

First of all, you should quote Apostol correctly: the statement about the existence of irrational numbers does not come finally, but rather at the beginning of section 1.10. This makes a difference, since the sentence may be interpreted as a pour parler. And indeed it is: Apostol assumes that $\mathbb{R}$ exists, so that the set of irrational numbers is just $\mathbb{R} \setminus \mathbb{Q}$. In my opinion, Apostol was thinking of very particular irrational numbers, like $\sqrt{2}$. You can define $\sqrt{2}$ as a supremum of a set of rational numbers (see for instance Rudin's Principles of mathematical analysis), and you need the axiom of completeness.

You did not write correctly the approximation property, but, to answer your question, Apostol's statement is true. The supremum may belong to $S$, so that a strict inequality would be wrong. And you are right when you write that sup=max when the sup belongs to your set.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.