Can the extended real number $+\infty$ be compared to transfinite numbers such as $\aleph_0$?

Question

If not, why not? If so, is ∞ greater than or less than $\aleph_0$?

Edit: the discussion in comments (including comments on a deleted answer) have made me think that the best way to put the issue is the following. Clearly, ∞ isn't a cardinal and $\aleph_0$ isn't an extended real number, so we can't use the usual order on either of these structures. Clearly also, an ad-hoc total order on the set $ℝ ∪ \{∞, \aleph_0\}$ says nothing about the extended reals or the cardinals. It seems like it might not be possible to define an order that extends the orders of both the extended reals and the cardinals in a "nice" way. But, given a suitable notion of "nice", how can we prove this?

Answer 1

Can the apple on my table be compared to the picture of an orange?

Well, by what property are you comparing them? Are they the same colour? No. Are they both fruits? Yes. Are they equally sweet? Meaningless question!

Similarly:

Can the $\infty$ from the affinely-extended reals be compared to $\aleph_0$?

By what property? Are they mathematical objects? Yes. Are they both greater than any natural number embedded into the respective structures (affinely-extended reals / ordinals)? Yes, but so what? Are they both the smallest such element in their respective structures. Yes, but again so what? Are they the same size? Meaningless question!

The whole point is that before you want to compare any two objects you have to specify your method of comparison. If you cannot specify that, then no comparison can be made. $\def\rr{\mathbb{R}}$

Clearly also, an ad-hoc total order on the set $\rr \cup \{\infty,\aleph_0 \}$ says nothing about the extended reals or the cardinals. It seems like it might not be possible to define an order that extends the orders of both the extended reals and the cardinals in a "nice" way. But, given a suitable notion of "nice", how can we prove this?

As I said in a comment, it's up to you to define "nice". But let me give you an example to explain why your question is too vague to be answerable. Note the ordinals up to $ω$ embed into the affinely extended real line with $w$ mapping to $\infty$. So what? That's my point. If this strikes you as very interesting, then by all means you have an answer. But if this seems to you to be meaningless because this mapping can't be extended to $ω+1$ or beyond unless the finite ordinals do not map to the natural numbers.

Now maybe you want to 'add points beyond $\infty$' in the affinely extended reals. Well you can, but then what? Already by adding the two infinities we lose the field properties of $\rr$, but at least we gain compactness. So what properties do you want to preserve in 'going further'? Maybe you like having an extended long line? It has totally no field properties, but is connected and sequentially compact (though not path-connected and not compact). And the ordinals up to $ω_1 = \aleph_1$ embed into the extended long line such that finite ordinals map to the natural numbers in the central copy of the real line.

Answer 2

I don't think we learn anything by trying to compare the $\infty$ of the extended real line to the cardinal number $\aleph_0$. But if you're interested in extending the real numbers into the realm of the transfinite, you may be interested in the surreal numbers. In the world of the surreals, weird-looking formulae like $$\frac{1}{1+\infty^2}$$ actually make sense, except they're usually written $$\frac{1}{1+\omega^2},$$ where $\omega$ is the least ordinal number that exceeds every natural number.

But $\omega$ is very different to the $\infty$ of the extended real line, e.g. we have $\omega+\omega > \omega$. Its an ordinal, not a cardinal.

A word of warning: I've got Conway's On Numbers And Games on my shelf, and its a pretty tough read, so don't expect the journey into the transfinite to be easy.