Is $\omega_0-1$ infinite?

Question

I have read in another answer Is infinity an odd or even number? that the $\omega_0$ is the "smallest infinity", but is $\omega_0-1$ not also infinite?

Answer 1

Nearly 8 years after he asked for it, but I'll finally add the answer Mark S requested in the comments on the OP.

The answer to the question depends on what is meant by two things:

• What does it mean to subtract $1$ from $\omega_0$?
• What does it mean to say something is "finite" or "infinite"?

The first question is part of a larger question of "where is this taking place"? $\omega_0$ is the least ordinal that is not a natural number, so the obvious choice is in the ordinal numbers, with ordinal arithmetic. And this is just what the other answers and comments before my own discussed. $\omega_0 - 1$ would then mean "the ordinal to which adding $1$ gives $\omega_0$". But there is still a problem: ordinal addition is not commutative. In general $a + b \ne b + a$, so the question arises of "adding $1$ to which side?" However, there is no ordinal $\alpha$ such that $\alpha + 1 = \omega_0$, but there is a unique ordinal $\beta$ such that $1 + \beta = \omega_0$. So if $\omega_0 - 1$ is to be given any meaning at all, it has to be that $\beta$, which is in fact $\omega_0$ itself.

There is a second arithmetic defined directly on ordinals, called the "natural addition" and "natural product". This turns out to be the same as the arithmetic they pick up as a subset of the Surreal numbers. However, there is no ordinal to which you can naturally add $1$ to get $\omega_0$.

John Conway and Donald Bluth's Surreal numbers include the ordinals as a subset. The Surreals themselves are a field (a Large field, as the Surreals are too big to be a set). They are the largest possible totally ordered field. So subtraction is defined, and $\omega - 1$ is a well-defined surreal number, but it is not an ordinal.

Thus in the Surreals we have $\omega - 1$. So we need to address the second question: what does it mean for a surreal number to be finite, or infinite? For ordinals, the answer is the obvious one. But the surreals also include numbers that are infinitesimally small, or involve the infinite in very complex ways, despite being less than finite numbers. I will ignore that and use the definition:

$\alpha$ is finite if and only if there are integers $n,m$ with $n < \alpha < m$.

In this case, $\omega - 1$ is not finite. If there were an integer $m$ with $\omega -1 < m$, then we would have $\omega < m+1$, which is also an integer. But no such integer exists. With this definition, "infinite $\pm$ finite = infinite" even in the Surreals.

Answer 2

It is clear what is meant by $5-1$ or $\mbox{one billion} - 1$, but to ask a question about $\omega_{0} - 1$ you first need to ask, "what do I mean by that"? How does one subtract $1$ from an object like $\omega_{0}$, which isn't quite a number? I'm not quite sure how much you know about ordinals so I'll try to keep this self-contained.

There is a notion of $+1$ that makes sense even for objects like $\omega_{0}$. $\omega_{0}$ is the infinite collection of all the natural numbers, and $\omega_{0} + 1$ is the infinite collection of all the natural numbers together with $w_{0}$. Put another way, $\omega_{0} + 1$ is a collection that has all the natural numbers together with the symbol $\omega_{0}$. You can subtract $\omega_{0} + 1$ by $1$, and indeed $(\omega_{0} + 1) -1 = \omega_{0}$.

When you ask about $\omega_{0} - 1$ you are asking for some quantity, say $\kappa$, for which $\kappa + 1 = \omega_{0}$. That is, you are asking for some infinite collection such that, with the inclusion of one larger element, you obtain the collection of the natural numbers. However, every finite collection of natural numbers is contained in a larger finite collection. There is no infinite collection coming right before $\omega_{0}$. In set theoretic language, this means that $\omega_{0}$ is a limit ordinal. As a result, you cannot subtract one from it.

Answer 3

The kicker, here, is that you need to define $\omega_0-1,$ in the first place. Typically, given a successor ordinal $\alpha,$ one might say that $\alpha-1$ is the order type obtained by removing the greatest element from $\alpha.$ However, $\omega_0$ has no greatest element (since it isn't a successor ordinal), so we can't define $\omega_0-1$ in this way.

Another approach one might take is to say that $\alpha-1$ should be the order type of the set of all non-greatest elements of $\alpha$ (which is equivalent to the other way among successor ordinals). In that case, since all elements of $\omega_0$ are non-greatest elements, then we have in fact that $\omega_0-1=\omega_0,$ counterintuitive though that may be. This latter definition is generally undesirable, though, because while in the first definition, $0-1$ is undefined (as far as order types go), the latter definition would make $0-1=0,$ which is wildly undesirable.

Answer 4

Surely it's an infinte number.The fact is that you have exactly $\omega_0=\omega_0-1 $. This can be proven knowing that $\omega_0$, the set of the natural numbers, is in bijection with $\omega_0-1$,that can be seen an the set of the naturals without the zero, via the function $n\mapsto n+1$,that is clearly bijective from $\omega_0$ to $\omega_0-1$. You can equally show this in the following way: for definition $\omega_0$ is the smallest of the infinte numbers.If $\omega_0-1$ was a finite number then also $\omega_0-1+1=\omega_0$ would be.Then $\omega_0-1$ is infinte and $\le\omega_0$.Since $\omega_0$is the smallest of the infinte numbers,it is necessary $\omega_0=\omega_0-1$.