Why we assume the least upper bound property to be true?

Question

I asked this question to my analysis teacher in the first lecture that why the least upper bound property which states that "any non-empty subset of real numbers which have an upper bound has a least upper bound" is true?

He replied it is an axiom and a fundamental property of real numbers. I know what axiom means we need to start somewhere to go somewhere just like Euclid's axioms but I am still curious about its acceptance.

Wikipedia states that there are several construction of real numbers which are equivalent to each other and this axiom is intuitive. But mathematicians who even work on the smallest details in the definition just why consider this to be an axiom without proving?

e.g. it is very intuitive to think that only a unique straight line in two dimensions can be drawn from two given points but the least upper bound property does not seem as intuitive as this.

Answer 1

We can demote the least upper bound property to a theorem in a fairly intuitive way.

Let's define what a real number is.

I'm not going to start from first-order real axioms because we run into a problem here, namely that there are many possible mathematical objects that we can construct that satisfy the real axioms (some of which are countable), for instance the computable reals or, depending on what kind of structure of the reals we want to express, the algebraic numbers might count (but would exclude $\pi$ and $e$).

Instead, let's take a Dedekind cut.

This is equivalent to saying that a real number is equivalent to all the rational numbers smaller than it:

$$ \text{the interpretation of $r$ is intuitively $\{ x : x \in \mathbb{Q} \land x < r \}$ } $$

This is a little circular though since we haven't defined what a real number is yet.

So, let's define a Dedekind cut as a set of rationals $X$ subject to the following rules:

1. Downward closed $ [\forall x \in X][\forall y \in \mathbb{Q}](y < x \to y \in X) $
2. No maximum element $[\not\exists x \in X][\forall y \in X](y \le x)$

Note that the inequality here is strict, so $\left(\frac{1}{2}\right)_\mathbb{Q}$ is NOT an element of $\left(\frac{1}{2}\right)_\mathbb{R}$.

And let's also rule out the empty set (corresponding to $-\infty$) and $\mathbb{Q}$ itself (corresponding to $\infty$).

Now, with that out of the way, suppose we have a bounded set of reals $\mathcal{F}$.

We stare at $\mathcal{F}$ real hard and notice that it is a set of sets of rationals.

We then compute $X = \cup \mathcal{F}$, the union of all these sets of rationals.

Note that $X$ is a real number and also that $X$ is greater than or equal to any element of $\mathcal{F}$ (as a real number). Additionally, any real number strictly smaller than $X$ will leave behind a rational that's present in one of the elements of $\mathcal{F}$.

Thus, as desired, every bounded set of real numbers has a least upper bound.

Answer 2

I suppose one could say people assume it, among many statements, for simplicity, in the rather specific sense that the other statements of interest that are regarded as 'equivalent' to it (and so, also regarded as 'true') are (or at least seem!) somewhat more complex:

• The l.u.b. property only makes reference to sets of reals (besides reals themselves), which are simply second-order objects (if we consider the reals themselves first-order)

• The compactness of the/any interval not only makes reference to all open covers, which are sets of sets of reals, ie, third-order objects, but also to 'finiteness', which is a notion that may not be available in setting up the theory

• The 'Archimedian property/axiom' invokes/makes explicit reference to objects of a certain sort/type, that is, natural numbers, that may seem to be just a harmless, 'obvious' constituent of the theory we're trying to set up, but are in fact a very different kind of beast (and I mean beast; model theorists speak of "tame" vs "wild" structures, take a look sometime)

• The nesting interval axiom/property/theorem makes reference to all (countable) sequences of intervals (satisfying some properties), but again, we may not have such (a notion of) an indexing available

• Bolzano-Weierstrass also uses sequences

• The connectedness of the/any interval involves, at least at first glance, "functions acting on second-order objects" (unions and intersections), hence third-order stuff, but maybe there is some clever way to work around that. Still, not so imediate, so not so appealing

• Maybe the 'Intermediate Value Theorem' could work, as as the interesting conditions to check regarding continuity can be expressed in first-order (the usual $\varepsilon$-$\delta$) terms, so maybe here specifically it's more of a psychological question of "We should not assume this seemingly more complicated thing, but rather prove it"

Certainly there are other statements/properties/axioms/whatnot that people do/could add to "ordered field" to try and characterize the reals, and any one of them should do the job; these are just the ones I recalled top of my head, but I hope it conveys the main idea that the l.u.b. property is just kinda easier to add to the theory and get it going. But then, if you're instead/still wondering "Why do we even need any of this stuff/these?", maybe take a look at the (greek-)constructible, the algebraic, and the computable numbers, and/or at the real-closed fields

Answer 3

The real numbers are defined as a non-empty set where + - * / < = > work as expected, and where ever non empty set with an upper bound has a least upper bound.

You then prove that there is indeed a set of real numbers, and that all such sets are basically identical. And then it turns out that the least upper bound makes lots of things work that don’t work with the rational numbers.

Answer 4

I interpret your question as asking for the intuition behind the least upper bound property.

Consider the integers: we have $0$, $1$, but intuitively we understand that there is a huge gap between these two: enter $1/2$.

Ok, now consider all the rationals. Are there any holes that remain? Yes! We can have a sequence of rationals growing larger and larger, remaining bounded, and still never converging to a rational (imagine a growing sequence of rationals whose limit is $\sqrt{2}$). We consider such a "limit point" (which doesn't exist yet!) of this sequence as a hole that we want to fill in. The least upper bound property is essentially imposing the condition: "if a sequence is growing and bounded, then the limit point must exist (no holes)". This property characterizes the completeness of the reals (no holes), an intuitive thing we want to achieve.

There are many more abstract spaces that have holes; the process of "completing" these spaces involves constructing the holes into existence, similar to how the holes in the rationals were constructed (forming the reals).

Answer 5

One way of making the least upper bound property intuitive is via nonstandard analysis. In axiomatic nonstandard analysis, there are two types of real numbers: standard and nonstandard. For example, an infinitesimal is a nonstandard number. An unlimited number, which is a number greater than every standard real number, is also nonstandard. A number is called limited if it is not unlimited, i.e., if it is smaller in absolute value than some standard real number.

It turns out that the least upper bound property is equivalent to the following property:

every limited number $x$ is infinitely close to a standard number $r$.

Such an $r$ is called the standard part of $x$. For example, the standard part of an infinitesimal is $0$. In other words, to get the standard part we have to discard an infinitesimal term.

Leibniz routinely used the procedure of discarding negligible terms in founding infinitesimal calculus. This is what is involved in passing from $2x+dx$ to $2x$ in the calculation of the derivative of $y=x^2$. So basically Leibniz relied on the least upper bound property (though of course he did not put it that way).

Answer 6

As far as I understand, the selection of axioms follows from geometric intuition. The reals are representing a line, and this must have no 'holes' if it represents an 'ideal' line in our heads(whatever is in quotations is not meant to be taken literally). We can look at this procedure in more generality as well, as done in Terrence Tao's Analysis-I text. We have rational numbers where we have defined distance between two rational numbers. If we look at Cauchy Sequences of Rational Numbers, or sequences of rational numbers such that the distance between any two of the terms of the sequence $(a_n)^{\infty}_{n=1}$ grows arbitrarily smaller, (Given any $\epsilon>0 \in \mathbb{Q}$, there exists $N_\epsilon$ such that $|a_n-a_m|<\epsilon, \forall n,m \geq N_\epsilon$), then if you look at the sets which are given as \begin{align*} E_n = [a_{N_{1/n}}-1/n,a_{N_{1/n}}+1/n] \end{align*} Then we clearly have $E_{n+1} \subseteq E_n, \forall n \in \mathbb{N}$, and moreover, the 'size' of $E_n$ grows smaller and smaller(of size 2/n), where each of them is still non empty. Now if we consider the set \begin{align*} \bigcap^{\infty}_{n=1} E_n \end{align*} you can show as an exercise that this set must either be empty, or a singleton set. If you try to draw this, we have a sequence of nested non empty sets such that they end either in a 'hole'(that is empty set) or a single element('The limit of the sequence'). $\textbf{Completeness}$ is the assertion that such kind of nested sets always end in a point, or such Cauchy sequences have a limit. Our intuition says this should happen for a geometric line but if we just take a look at the rationals, it turns out this is not the case. So what do we do? It is quite simple- we literally fill those holes in by saying that any sequence which tends to the 'same' hole is collected in a box and called a number- and this is what we call the real numbers. The least upper bound property can be shown to be equivalent to this property.