Is there a way of defining infinity so that it both meets the criterion of being a number as well as being greater than any real number?

Question

This number would not be equivalent to the concept of infinity, but would be a number analogous to that concept. I'd also be interested to how the different ways of doing this (I'm assuming there are many), how they differ, and what their different applications are.

So this version of infinity would behave like this:

$\infty$ > X for all X $\in$ $\mathbb{R}$

$\infty$ + 2 > $\infty$ + 1

2 > 1

Rather than:

$\infty$ is the largest number

$\infty$ + 2 = $\infty$ + 1

2 = 1

Answer 1

There are indeed lots of ways to do this, although they may not quite paint the picture you expect.

At the simplest level, assuming that you want a number system in which addition, subtraction, multiplication, and division work "as we expect," you're just looking for non-Archimedean ordered fields which contain $\mathbb{R}$. There are indeed lots of these. For example, consider the set of rational functions in a single variable $x$ with coefficients from $\mathbb{R}$, modulo equality almost everywhere (so that we identify e.g. $x$ with $x^2\over x$). This is a field, and carries a natural ordering compatible with the field structure: set $f<g$ iff $\lim_{x\rightarrow\infty}(g(x)-f(x))>0$. It's easy to check that this is in fact a non-Archimedean field. (In fact, note that any ordered field containing $\mathbb{R}$ must be non-Archimedean!)

• A powerful result here is the compactness theorem from logic. This implies that, in a precise sense, whenever I have a "reasonably concrete" set of algebraic properties describing some (infinite) structure there are other structures satisfying those same properties which are "as big as I want" (e.g. cardinality $2^{2^{2^{2^{\aleph_0}}}}$ instead of $\vert\mathbb{R}\vert=2^{\aleph_0}$). That's overkill here, but it's a good thing to keep in mind for down the road. (Note that this, combined with the previous paragraph, means that Archimedean-ness is not a "reasonably concrete" property in the relevant sense!)

We can look further for non-Archimedean fields satisfying additional special properties. A hyperreal field, for example, is a non-Archimedean ordered field which shares all "reasonably-definable" properties with $\mathbb{R}$ in a precise sense, while the surreal numbers form a particular (proper-class-sized) non-Archimedean ordered field which is "maximal" in an appropriate sense.

In all such cases however there is not a single "distinguished" infinite element; instead, there are lots of infinite elements, with no one being particularly special. So this cuts against the idea of $\infty$ as a meaningful symbol denoting a single element.

Answer 2

As Noah says there are a bunch of inequivalent ways to do this depending on what properties of ordinary arithmetic you want to preserve. Here's what happens when you try to add a single infinite element.

Starting with the real numbers $\mathbb{R}$ (or the integers or whatever else) you can add a new element $\infty$ and declare that $x + \infty = \infty$ for any $x$ (including $\infty$). This works fine (considering only addition for now, no multiplication) except that you lose the ability to subtract; this addition is no longer cancellative, meaning that $x + y = x + z$ no longer implies $y = z$. Formally (ignore this if you don't care), addition only makes $\mathbb{R} \cup \{ \infty \}$ a monoid rather than a group, and $\infty$ becomes an absorbing element. But we can still define an order in which $x \le \infty$ for all $x \in \mathbb{R}$ and this order still has the property that if $x \le y$ then $x + z \le y + z$; this means we still have an ordered monoid.

Now suppose we want to get multiplication in the game. Probably we should ask that $x \cdot \infty = \infty$ for $x > 0$ and in particular for $x = \infty$. But we need to decide what $0 \cdot \infty$ is. If we insist that multiplication continue to distribute over addition then we need

$$0 \cdot \infty = 0 \cdot (\infty + \infty) = 0 \cdot \infty + 0 \cdot \infty$$

which means the only possible options are that $0 \cdot \infty = 0$ or $\infty$. If we pick $\infty$ then we lose the familiar fact that $0$ times anything is zero. I don't think $0$ is an entirely satisfactory option but let's stick with it for now.

Now we need to decide what $(-1) \cdot \infty$ is. Again, if we insist that multiplication continue to distribute over addition, this requires

$$0 = 0 \cdot \infty = (1 - 1) \cdot \infty = \infty + (-1) \cdot \infty.$$

But with the way we've defined addition this is impossible; there is no element which, when added to $\infty$, produces $0$. The general problem here is that distributivity forces $-1$ to act as the additive inverse, and we already gave that up. So we have a couple different choices here:

• We could give up distributivity. That would be terrible.
• We could add a new element $-\infty$ and declare that $(-1) \cdot \infty = - \infty$ and that $\infty + (-\infty) = 0$. But this breaks associativity: for example, we would have $0 = \infty + (-\infty) = (1 + \infty) + (-\infty) = 1 + (\infty + (-\infty)) = 1$. The lesson is that when we gave up subtraction we really gave up subtraction and we shouldn't try to get it back. Subtraction is gone forever.
• We could give up negative numbers. This is actually a perfectly fine thing to do: the non-negative reals $\mathbb{R}_{\ge 0} \cup \{ \infty \}$ together with infinity, addition, and multiplication as defined here is a perfectly well-behaved ordered semiring (which basically means we can add, multiply, and compare $\le$, but we can't subtract), and the subring consisting of the non-negative integers together with $\infty$ is exactly the ordered semiring given by the at-most-countable cardinal numbers.

We have to give up something, though.