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What is the last term of the set of natural numbers $\Bbb{N}$? You may say one does not exist, because whatever number $n\in\Bbb{N}$ you CHOOSE, you can always find a natural number greater than that.

What if I were to define the last term of $\Bbb{N}$ as a natural number which is larger than any natural number you can choose...that the last term does exist, but we simply can't choose it? Would this definition be valid?

Allowing such a definition would validate a lot of things that are intuitively true. For example, any number we select in $(0,1)$ can be shown to belong to an open set that is a subset of $(0,1)$. However, intuitively, we feel that this might be not true for EVERY number in $(0,1)$. Because that last number, just before $1$, would not be contained in any such open set. However, for any $r\in(0,1)$ we choose, we can create such an open interval within $(0,1)$. So maybe the greatest element of $(0,1)$ (and NOT the least upper bound) is an element we can't choose, but define, such that its existence is implied....?

There seems to be a bit of a chasm between my intuition and some mathematical facts before me. This is a most amaterish effort on my part to plug that. Does something like this already exist?

Thanks in advance!

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closed as not a real question by Andres Caicedo, Gottfried Helms, Brian Rushton, Hagen von Eitzen, MJD Jun 21 '13 at 6:08

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I'd recommend reading about the surreal numbers. The question you ask about $(0,1)$'s "greatest number" may be answered by looking at the surreal number $\alpha <(0,1)|1>$, which has the property that for any $x \in (0,1)$ $x < \alpha < 1$. On the other hand, the "least upper bound" for $\mathbb{N}$ would be $\omega =<\mathbb{N}|\emptyset>$ –  andybenji Jun 21 '13 at 4:26
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I think the surreal numbers is a big step and will feel like sorcery if he begins with that. Understanding the notions of maximum, maximal elements, supremum and partial orders would be a way to begin ; those things are nasty generalizations of what happens over the reals, so it is better to understand what happens there first. –  Patrick Da Silva Jun 21 '13 at 4:36
    
Any $x \in (0, 1)$ is already in the open set $(0, 1) \subseteq (0, 1)$. What are you even trying to say? –  Vectk Jun 21 '13 at 5:33
    
@andybenji- Surreal numbers address my doubts perfectly. Thanks! –  Ayush Khaitan Jun 21 '13 at 5:36
    
@Ink- I suppose a better formulation of the question can be any $x\in (0,1)$ is contained in an open interval $(a,b)$, where $a,b\in (0,1)$. –  Ayush Khaitan Jun 21 '13 at 5:37

3 Answers 3

You seem to assume that there will always "be a last number" or that "every ordered set admits a maximum", which is not true (a maximum would be an element which is bigger than everyone else). In the case of $]0,1[$, it doesn't have a maximum, nor does it have a supremum (a supremum is the smallest possible upper bound), but we can see that if we look at it as a subset of $\mathbb R$, then $1$ would be the 'smallest possible upperbound for the set $]0,1[$, and that is what we call the supremum of $]0,1[$ in $\mathbb R$.

Perhaps other examples of partial orders would help put your brain in the general setting and understand better what happens in the case of $\mathbb R$. In my case it helped a lot.

A set $X$ together with a binary relation $\le$ is said to be partially ordered by $\le$ if the following hold : $\forall a,b,c \in X$

  • $a \le a$
  • $a \le b$ and $b \le a$ implies $a = b$
  • $a \le b$ and $b \le c$ implies $a \le c$.

It is possible that $a \not\le b$ and $b \not\le a$ (i.e. those axioms do not imply that last statement). If in addition to those properties we ask that $a \le b $ or $b \le a$ for every $a,b \in X$, the partially ordered set is said to be totally ordered.

As an example, consider the set of integers $\mathbb N$ ; for $a,b \in \mathbb N$, say that $a \le b$ if $a$ divides $b$. Therefore $2 \le 4$, $2 \le 26$ and $13 \le 39$ ; this seems to coincide with the usual partial order, but be careful ; we have $2 \not\le 3$ and $3 \not\le 2$!

Now with a bit more definitions :

  • An element $a \in X$ is said to be a maximal element if there is no $b \in X$ such that $a \le b$ with $a \neq b$.
  • An element $a \in X$ is said to be a maximum for $X$ if for every $b \in X$, $a \le b$.

Note that these two definitions may seem similar (they are in the case of totally ordered sets), but in the case of partial orders they are not! Consider $\{2,4,7,8\}$ partially ordered by division (i.e. $a \le b$ if $a \, | \, b$). Then $7$ and $8$ are maximal elements because $7$ and $8$ divides no element besides themselves. But this set admits no maximum! In other words, there is no "largest element" in this set.

So my point is you must not take for granted that maximums exists. Now let's talk about supremums.

  • Let $A \subseteq X$ where $X$ is partially ordered by $\le$. We say that $A$ admits $b \in X$ as a supremum if the two following conditions hold :

    1. For every $a \in A$, $a \le b$ ;
    2. For every $c \in X$ such that $c$ satisfies 1., (i.e. $a \le c$ for all $a \in A$), then $b \le c$.

If this supremum exists, we will denote $b = \sup A$. (Note that by 2., if two elements $b$ and $b'$ were to be supremums, we would have $b \le b'$ and $b' \le b$, hence $b=b'$, so we can speak of THE supremum).

Let us consider my partial order as before $\mathbb N$ with partial order given by dvision. Then I claim that $\sup \{2,4,7,8\} = 56$. Indeed, if $a \in \mathbb N$ is such that $2,4,7,8 \, | \, a$, then $56 \, | \, a$, which proves 2. ; but $2, 4, 7$ and $8$ divides $56$, so this makes $56$ satisfy 1. and $56 = \sup \{2,4,7,8\}$.

What if $A = \{ 2,4,8,16,32,\dots,2^n,\dots,\}$, the set of all powers of $2$? This set has no maximal element because for every $2^n \in A$, $2^n \le 2^{n+1}$. It has also no upper bound, because a positive integer which is divisible by all powers of $2$ does not exist. Thus we say that this set admits no maximum, no maximal element, no supremum. We need to accept that our definitions lead to such a thing.

Now back to your example. $\mathbb R$ is a totally ordered set, so any two numbers are comparable. $\mathbb R$ also has the following amazing property ; any set bounded above admits a supremum! So for instance, the supremum of $]0,1[$ would be $1$ ; every number strictly between $0$ and $1$ is smaller than $1$, and any number which is bigger than any number in $]0,1[$ is greater or equal to $1$. Therefore $\sup \,]0,1[ = 1$. It is not a problem that the supremum of $A$ is not in $A$ ; it happens.

What about subsets who are not bounded above? Like for instance, what about $[0,\infty[ \subseteq \mathbb R$? This set is not bounded above, hence cannot have a supremum, because no element of $\mathbb R$ bounds by above every positive real number.

If you have every heard about the completion of the real line, it is useful for instance in measure theory to consider $\overline{\mathbb R} = \mathbb R \cup \{-\infty, + \infty\}$ so that every subset of $\overline{\mathbb R}$ is bounded below and above ; the subset $[0,\infty[$ would therefore be bounded above by $\infty$. If you are ready to accept what I just said then it would be the best approximation to your wild dreams that are accepted by the mathematical community.

Hope that helps,

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Erm....however little understanding I may have of math, I already know all this. And although I may be mistaken, this post has little relevance with what the question reallly is. –  Ayush Khaitan Jun 21 '13 at 4:45
    
Thanks for your time and patience. But the reason why I think this post has little to do with the question is you have CHOSEN an element from the set of even numbers, and said that a greater number than that exists. That is the sole bone of contention. What if we couldn't choose some numbers in an infinite set? –  Ayush Khaitan Jun 21 '13 at 4:47
    
I think you're dismissing this question completely out of hand. Which it may be. But you haven't directly addressed my concerns. I'm well aware of the definitions of maximal elements and supremums, etc. But as of now, I don't think they address the question. Thanks again –  Ayush Khaitan Jun 21 '13 at 4:49
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Okay, before dismissing my answer as completely irrelevant, I am sorry if I wasted 30 minutes trying to explain you what I thought you didn't understand. (I'm trying to say : don't be rude.) I have seen people before being confused about those stories of supremums and infimums and I helped them with a similar kind of answer so I thought it would help you too. If you are really in for 'finding holes in the set of real numbers' and you think you could understand, then perhaps what you should do is read about super-real numbers, hyper-real numbers or simply the surreal numbers. –  Patrick Da Silva Jun 21 '13 at 4:57
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These add a notion of 'infinitesimal' (i.e. elements that are strictly smaller than any positive real number, yet strictly greater than zero, hence are in a "hole between $\{0\}$ and $]0,1]$") and the surreal numbers is the ordered field which has "no more holes" in some sense (it always depends on what sense you say it, because I mean, the real numbers has no holes in the sense of supremums, but we can always dig deeper, can we?). –  Patrick Da Silva Jun 21 '13 at 4:58

What if I were to define the last term of $\Bbb N$ as a natural number which is larger than any natural number you can choose...that the last term does exist, but we simply can't choose it? Would this definition be valid?

In order to make this a legitimate definition of anything, you would at the very least have to define exactly what you mean by choosing a number. To me the statements we can choose an element of $A$ and $A\ne\varnothing$ are synonymous, so $\{a\in A:\text{we can choose }a\}=A$. From this point of view your attempted definition is incoherent: in effect it says that a certain subsetof $\Bbb N$, namely, the set of natural numbers that we cannot choose, is both empty and non-empty.

Even supposing that you could define choosable in a rigorous, meaningful fashion, you would not be defining ‘the last term of $\Bbb N$’: $\Bbb N$ is a specific mathematical object that simply does not have (in its customary order) a last element. This is explicit in any axiomatization of $\Bbb N$ that I’ve seen and in the (almost?) universally shared intuition about the natural numbers that these axiomatizations formalize.

There are philosophical approaches to mathematics that involve notions that are, I think, somewhat similar to your notion of choosable and that can be given rigorous expression; then come under the heading of constructivist mathematics. I frankly have very little use for them and can’t tell you much about them, but you may find them of interest.

Allowing such a definition would validate a lot of things that are intuitively true. For example, any number we select in $(0,1)$ can be shown to belong to an open set that is a subset of $(0,1)$. However, intuitively, we feel that this might be not true for EVERY number in $(0,1)$. Because that last number, just before 1, would not be contained in any such open set.

That is an intuition that I’ve never shared, because it is so obviously false: given any $x\in(0,1)$, $\frac12(x+1)$ is clearly a larger element of $(0,1)$, halfway between $x$ and $1$, so $(0,1)$ plainly has no last element.

In terms of doing ordinary mathematics (as distinct from worrying about foundational and philosophical issues) I’d focus on trying to modify your intuition to bring it into line with the mathematics as it’s (almost) universally done: I think that trying to create from whole cloth a philosophical plug or bridge to fill or span the gap between intuition and mathematics will take you into very murky waters very quickly and is more likely to get in the way of understanding the mathematics than to help you to understand it.

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If you have a mathematical definition of "choose", that would help greatly.

If "choosing" something means "let a variable equal that something", then what is the purpose of an object we cannot choose? We cannot refer to it at all. Depending on your definition of "exists" ($\exists$), you might not even be able say it exists. I would argue that definition of "choose" is useless.

However, if by "choose" you mean "define", then yes, there are such numbers we cannot define. The number of formulae we can write is countably infinite, but the reals are uncountable, and so we cannot define most of them. I believe one can assign a variable to them, although one will not know much about what they have chosen. So this might be valid: Let $S$ be the set of real numbers not definable in (however one mathematically defines a language). Let $x$ be an arbitrary member of $S$. We can show $x$ isn't algebraic, I suppose? Not sure what can be done that is practical though.

If "choose" means "construct", then yes, there are numbers we cannot construct. But we can still define them and assign to them a variable. For example, we are definitely allowed to state something like: "By the Intermediate Value Theorem, there exists some $x$ such that $f(x) = 4$", and we are allowed to manipulate that $x$, despite not being able to construct such an $x$ explicitly.

The second definition does not apply to either number you suggest, as we can define both of those. Even under the third, we can show that such numbers do not exist in $\mathbb{N}$ and $\mathbb{R}$, respectively.

(small question, what exactly is the difference between constructible and definable? Is constructible the same as computable?)

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This doesn't feel like an answer to me but more like a long comment. –  Patrick Da Silva Jun 21 '13 at 5:34
    
Sorry, edited to be more answer-y. –  Henry Swanson Jun 21 '13 at 5:44

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