224
I give back what I recieve just as the law requires
(lemmy.dbzer0.com)
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 12 Aug 2023
224 points (100.0% liked)
196
16505 readers
2104 users here now
Be sure to follow the rule before you head out.
Rule: You must post before you leave.
founded 1 year ago
MODERATORS
We have sorta the same problem with imaginary numbers, and I remember some programmable calculators can process complex numbers using symbolic representation (which happens to work similarly to Cartesian coordinates, so that's convenient)
But from what I remember any infinity bigger than counting numbers (say the set of real numbers) cannot be differentiated from each other, so we don't have established rules.
To be fair, I last tinkered with infinities in the aughts and then as a hobbyist. The Grand Hilbert Hotel can accomodate more compound infinities and still retain perfect utilization since the last time I visited.
https://en.wikipedia.org/wiki/Continuum_hypothesis
Hmm. Frankly speaking I always assumed Mathematicians had more of an idea about infinities I mean why even have indices if you don't have an inductive rule to descr... oh wait never mind.
That said the reals aren't countable yet we have perfectly reasonable ways to deal with them symbolically, even compute with them, represent ~~every single one of~~^1^ them in finite space, it's just when you want to compare or output them with infinite precision that you might have to wait for eternity. But who needs infinite precision anyway, arbitrary precision is plenty.
^1^ On second thought, after diagonalisation, no we don't. Or we do because there's some magic going on with included transcendental constants that break through that do I look like a numerologist.
there actually is a way to represent the reals with full generality in homotopy type theory -- work is still on-going to implement it in a real programming language/prove type checking is decidable, but the theory is already in place -- via Cauchy sequences.
it's more than convenient, it's isomorphism!
I don't understand what you think the problem is. What do you mean infinities can't be differentiated from each other? Infinite cardinals are by definition equivalence classes of sets that can be placed in bijection with one another. You can compare one cardinal to another by showing there exists an injection from a representative set of the first one into a representative for the other. You can show equality by showing there is an injection both ways (Cantor-Schroder-Bernstein theorem) or otherwise producing a bijection explicitly. Infinite ordinals may as well be copies of the natural numbers indexed by infinite cardinals, so of course we can distinguish them too.
So far AFAIK we have two kinds of infinity: Those that can be accommodated at the Grand Hilbert (e.g. integers, fractions, etc.) and those that cannot (set of irrational numbers, set of curves, set of polytopes, etc.) This was why we had to differentiate orders of infinity, e.g. ℵ₀ (The Grand Hilbert set), ℵ₁ (the irrational set, the real set), ℵ₂ (???), ℵ₃ (?????), ℵₙ (??!??????????)
For values of infinity that are in higher orders than ℵ₀, we can only tell if they're equal to ℵ₁ or undetermined, which means their infinity size is ℵ₁ or greater, but still unknown.
Unless someone did some Nobel prize worthy work in mathematics that I haven't heard about which is quite possible.
No, that's definitely not true. As I said, infinite cardinals (like the cardinality of the naturals ℵ₀) are defined to be equivalence classes of sets that can be placed in bijection with one another. Whenever you have infinite sets that can't be placed in bijection, they represent different cardinals. The set of functions f : X --> X has cardinality 2^X too, so e.g. there are more real-valued functions of real numbers than there are real numbers. You can use this technique to get an infinite sequence of distinct cardinals (via Cantor's theorem, which has a simple constructive proof). And once you have all of those, you can take their (infinite) union to get yet another greater cardinal, and continue that way. There are in fact more cardinalities that can be obtained in this way than we could fit into a set-- the (infinite) number of infinite cardinals is too big to be an infinite cardinal.
You might be thinking of the generalized continuum hypothesis that says that there are no more cardinal numbers in between the cardinalities of power sets, i.e. that ℵ₁ = 2^(ℵ₀), ℵ₂ = 2^(ℵ₁), and so on.
oh this is a neat argument I'd never encountered before. I was also under the impression that we hadn't proved there were infinities with cardinality greater than ℵ₁.
How/why would you simultaneously hold this belief and reference the HoTT book
because I was misinformed?
What are you doing with the HoTT book if you have never heard of Cantor's theorem??? You must understand there's a minimum of several years of intensive study in between these two things
self-study. it's been a decade since I was in school and kept encountering references to it, so I've been working through a lecture series and the book.
It's quite possible that what I'm encountering is the the momentary failure to understand Cantor's theorem, or rather the mechanism it uses to enumerate the innumerable. So my math may just be old.
Cantor's theorem says the power set of X has a strictly larger cardinality than X.
When |X| is a natural number, the power set of X has cardinality 2^(|X|), since you can think of an element of the power set as a choice, for each element of X, of "in the subset" vs "not in the subset." Hence the notation 2^X for the power set of X.
Cantor's theorem applies to all sets, not just finite ones. You can show this with a simple argument. Let X be a set and suppose there is a bijection f : X -> 2^(X). Let D be the set { x in X : x is not in f(x) }. (The fact that this is well defined is given by the comprehension axiom of ZFC, so we aren't running into a Russell's paradox issue.) Since f is a bijection, there is an element y of X so that f(y) = D. Now either:
y is in D. But then by definition y is not in f(y) = D, a contradiction.
y is not in D. But then by definition, since y is not in f(y), y is in D.
Thus, there cannot exist such a bijection f, and |2^(X)| != |X|. It's easy enough to show that the inequality only goes one way, i.e. |2^(X)| > |X|.