Is there anything wrong with my interpretation of these exercices about Dedekind left set?

Question

There seems to be serious misunderstanding between me and my textbook ("Classic Set Theory. A guided independent study" by Derek Golderi).

I must provide you with definition of Dedekind left set because it's important to understand the exercises:

FACEPALM. So far so bad. "A real number is Dedekind left set" doesn't make sense to me. A number and a set are two different things. Yes, there can be a set whose the only element is a number, but it does't make the set itself a number.

There are two exercies that have nonsensical places too (each one has a solution). Let's begin with the second one because it's simpler:

ME: Of course it's not a real number, it's not even a number, it's a SET!

Now let's look at the first exercise:

Definition of set q is absurd. It basically says "We have set q that consists of rational numbers and each its rational number has following property: namely any rational number that is less than any rational number". But there is no such number! It could make sense if we were comparing a set of negative numbers to a set of positive numbers, after all any negative number is less than any positive number. But we don't compare sets of negative and positive numbers, so in this case we get empty set.

Let's check the solution:

1.This is not clear what means. Does it mean "q+1 isn't less than q when assumed that q is a rational number"? If yes, then why don't we just write it down as "q+1 is more than q when assumed that q is a rational number"? It's more straightforward. Besides, if we adopt the interpretation it's unclear why should we conclude that q is a proper subset of all rationals.

2.I'm not sure how to interpret and why should it mean that set q can't be empty. My best guess is that it means "For any rational number there is always smaller rational number exists. Thus the set is NOT empty". I agree with the premise, but I don't see how the conclusion follows from it.

3.The rest of the solution is unclear for me too, but I guess I'll be able to understand it if I will have clear understanding of already mentioned places that confuse me.

Answer 1

I think you have to get used to the idea that a thing in mathematics is what we define it to be. In general there could be multiple definitions which don't have any conflict with each other. And one of these definitions could be more popular to become common knowledge. But in any case if one encounters an unfamiliar definition one should try to accept it and see where it leads us.

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Here you encounter the definition of "real number" as some sort of a subset of the set of rationals. If you find this definition as bit disturbing then understand that "real numbers" are difficult to define and most high school textbooks conveniently avoid defining them.

Another confusion is the use of same letter to denote a real number and a rational number. Observe that rational numbers in your book are denoted by lowercase Roman symbols $a, b, \dots, p, q, \dots$ whereas the same symbols in bold like $\textbf{p, q} $ represent Dedekind left sets ie real numbers. Once you notice this carefully most of the doubts about the two problems will be gone.

Let's start with $$\textbf{q} =\{p\in\mathbb {Q} \mid p<_{\mathbb{Q}} q\} $$ Here $q$ is a specific rational number (you can think of some example like $q=1$ in your mind if that helps). And $\textbf{q} $ is a specific subset of $\mathbb{Q} $ which depends on rational number $q$. In simple English $\textbf{q}$ is the set of all rationals which are less than $q$ (for example you might consider the set of all rationals less than $1$).

To show that this set $\textbf{q} $ is Dedekind left set just check that it satisfies the three properties given in the definition of Dedekind left set.

It is clearly not empty because there are many rationals less than $q$ (in particular $q-1$ is one of them) and they all belong to set $\textbf{q}$. Also it is a proper subset of $\mathbb {Q} $ because there are many rationals which don't belong to it (precisely $q$ and all greater rationals don't belong to $\textbf{q}$).

Next property of Dedekind left set when translated into English reads like "if a rational number belongs to a Dedekind left set then all the smaller rationals also belong to it". If some rational $x\in\textbf{q}$ then by definition of set $\textbf{q} $ we have $x<q$ and clearly all rationals less than $x$ are also smaller than $q$ so that all rationals smaller than $x$ also belong to $\textbf{q} $. Thus the second property is also verified for $\textbf{q} $.

The final property of Dedekind left set needs to be verified for $\textbf{q} $ and it reads that "there is no greatest member in $\textbf{q} $" ie "if a rational $x$ lies in $\textbf{q} $ then there is another greater rational $y$ which also lies in $\textbf{q}$". Let $x\in\textbf{q}$ then we have $x<q$ and the rational number $y=(x+q) /2$ is such that $x<y<q$. Since $y<q$ we have $y\in\textbf{q} $ and thus we have found a $y$ greater than $x$ which lies in $\textbf{q}$.

Answer 2

The problem is: How do we define/construct the real numbers?

We can define/construct the rational numbers by iteratively adding, subtracting the number one and dividing the quantities we get. But that won't give us the irrationals.

Now we can't just say "the irrationals are everything else" because that is not a definition.

So if we can't define the irrationals why do we need them? Well, because in comparing sizes of the rationals we can "get as close as we like" between them. This implies that we desperately want there to be a "continuum" that if we go between two numbers we can pass through every value between.

The rationals fail this. And we can't just declare something exists without knowing what it is. So is there any way we define this continuum using the concept of rational numbers?

Well, the rational numbers get close and for every desired value in this continuum we can get an infinite number of rational numbers that get close to it with infinite precision by considering this infinite set of rational numbers.

In other words. For every value in the continuum, there is a unique set of all the rational numbers that are less than it that we can uniquely associate with this value.

And that is the answer to our problem. We will define/construct the real numbers by these unique sets of rational numbers by associating each real number with a unique set of rational numbers. (and vice versa.)

Now, you might quibble, but those sets are associated with the real numbers. The are not the real numbers themselves. Point taken. But I claim it does not matter; it is a one to one association and thus a unique defining. As these are abstractions an association (provided it always exists and is always unique and consistant) is all there is.

Hence: A real number is defined by the unique set of all rational numbers that are less than it.

These sets are called Dedekind left sets and as they are uniquely associated we define a "real number" as being the set.

==== old answer below ====

The book is replacing everything you know about numbers with something different.

The problem is if we use the concept of numbers that we grew up with, CONTWGUW, then we can define the rationals but we can not define the reals. We can't just say the reals are "everything else" or with the "holes filled in".

So the book is doing something radical. It is defining a mathematical system that is not CONTWGUW but absolutely equivalent to it and it is going to use that instead. Forever and always. It will never use CONTWGUW again.

We start by defining the rationals as we always have and the arithmetic from this. But because we are going to abandon it entirely we want to indicate it is the old arithmetic, we will note it with the $_\mathbb Q$ symbol.

For example $q +_\mathbb Q 1$ will mean $q$ plus $1$ with the old arithmetic. And $q <_{\mathbb Q} r$ means $q < r$ in the old arithmetic.

So the book defines a type of set. A Dedekind left set. The definition makes sense. And by definition we are going to call a Dedekind left set a "real nummber". This is fine! We could have called a Dedekind left set a "pickle sandwich" if we wanted to. It does not matter that "real number" means something different in CONTWGUW. We are not ever going to use CONTWGUW ever again.

So. Is $r_* = \{q\in \mathbb Q| q\le \frac 12\}$ a 'real number' or a Dedekind left set.

No, it is not because it has a maximum element, $\frac 12$ and a 'real number' (which is a type of set) doe not have a maximum element.

Now if $q \in \mathbb Q$ and $q_* = \{p \in \mathbb Q|p < q\}$, is the set $q_*$ a 'real number'?

Your first mistake is you describe $q_*$ as "We have set q that consists of rational numbers and each its rational number has following property: namely any rational number that is less than any rational number".

It is not the set of any rational number less than any other; it is the set of all rational numbers less than $q$ which is a specific rational number.

"This is not clear what $q +{\mathbb Q} _1 \not < q$ means. Does it mean "q+1 isn't less than q when assumed that q is a rational number"?

Yes, it means exactly that.

"If yes, then why don't we just write it down as "q+1 is more than q when assumed that q is a rational number"? It's more straightforward."

Why is that more straightforward? We want to show that $q+1 \not \in q_*$ and $q+1 \in q_* \iff q+1 < q$. So we want to show it is not true that $q+1 < q$. i.e. we want to show that $q+1 \not < q$.

"Besides, if we adopt the interpretation it's unclear why should we conclude that $q_*$ is a proper subset of all rationals. "

Um, $q_* = \{p\in \mathbb Q| p< q\} \subset \{p \in \mathbb Q\} = \mathbb Q$. So it is a subset. We just need to show it is a proper subset. It is not empty because $q -1 \in \mathbb Q$. And it is not all of $\mathbb Q$ because $q + 1\not \in \mathbb Q$. So it is a proper subset.

"I'm not sure how to interpret $q-_{\mathbb Q}1 <_{\mathbb Q} q$ here and why should it mean that set $q_*$ can't be empty. My best guess is that it means "For any rational number there is always smaller rational number exists. Thus the set is NOT empty". I agree with the premise, but I don't see how the conclusion follows from it."

If means $q - 1 < q$ . Thus $q-1 \in \{p\in \mathbb Q|p < p\}=q_*$. Thus $q_*$ is not empty.

The two remaining things we have to do are:

ii) Show that if $p \in q_*$ and $r < p$ ($r$ is rational) then $r \in q_*$.

This is clear. If $p \in q_*$ then $p$ is rational and $p < q$. And if $r < p$ and $p < q$ then $r < q$ and therefore $r\in q_*$.

iii) Sow that $q_*$ has no maximum element.

If can't. If $r \in q_*$ then $r$ can not be maximal. Because $r< \frac {r+q}{2}< q$ . So $\frac {r+q}2 \in q_*$.

So All the conditions of a Dedekind left set are fullfilled.

So $q_*$ is a Dedekind left set. So it is what we are calling a 'real number'.

Answer 3

In wikipedia link Zermelo–Fraenkel set theory you will find this sentence:

ZFC is intended to formalize a single primitive notion, that of a hereditary well-founded set, so that all entities in the universe of discourse are such sets.

And you can drill down to Hereditary set if you want to be further educated (entertained!).

If we ever get into trouble we can dig down into this abstract ZFC formalism.

Your book is going to construct the real numbers within this set theoretic framework using a model (Dedekind left sets). But practically, you will soon be able to work with and understand numbers and not even be thinking of this model - you'll have your own 'model' in your head.