Bad pun about uncountable infinity

[jack]

Q. OK, so what about uncountable infinities in magic?

What about it?

Q. Does it make infinite combos less degenerate? Does it even make sense?

Well, you may need some tweaks to the rules, because a lot of them are predicted on the idea of things happening one after another. But suppose, eg. you just go round an infinite loop ω₁ times.

Q. Well, that was boring.

OK, ok. Suppose you can't do that. Try this. Doubling season says "Whenever you put a counter into play, instead put two of them into play." Suppose you get an infinite number of doubling seasons into play (eg. jumping through a few hoops and an infinite amount of mana to put a copy of it into play an infinite number of times).

Then put a token into play. I think the infinite number of doubling seasons produces an uncountable number of new counters. (Either an uncountable number of new creatures, or almost cooler, one creature with enough +1/+1 counters on that it has power and toughness ℵ₁/ℵ₁ :))

Think about this either as 2^ω counters, or by counting the counters individually: number the first counter 0, and the counter produced by the first double 0.1 and the counters produced by the second double 0.01 and 0.11, and so on.

Q. Right... Is that rigorous?

I think so... You need either to be comfortable with an infinite stack, (the order things come off it might be a problem?) or to have events happen "in parallel" when it's obvious what's going to happen.

Q. Is it interesting?

I don't know. Maybe. Probably not. It does mean that if you gain uncountably infinite life, your opponent has to jump through some extra hoop, it's not enough just to have an infinite combo. You don't just need a way of gaining infinite cards, infinite mana, and infinite damage, you need an infinite number of copies of some doubling effect as well.

Q. So, specific cards?

Aforementioned doubling season, together with a card that makes it a creature (opalescence) -- or anything else if you can copy it -- and a reusable card that puts a copy of target creature or artifact (kiki-jiki "tap to put into aply a copy of target creature" plus a way to untap it, mirrorweave "all creatures become a copy of target creature" plus infinite creature tokens, spitting image "pay mana and discard a land to repeatedly put a copy of target creature into play").

Isn't there some red enchantment which does twice as much damage?

Q. I ask the questions here.

That's not a question.

Q. I ask the questions here. Right, asshole?

Comments

[alextfish.livejournal.com]

Yay, uncountability in Magic. It's quite hard to do (assuming you disallow going through countable steps ω₁ times). PeterTaylor came up with the following dubious combo: Generate a (countably) infinite number of creature tokens, with a Juniper Order Ranger in play to give them all +1/+1 counters. Make them all artifacts, with Mycosynth Lattice. Also, have Gleemax in play, as well as some indestructible artifact with at least one counter on it. Cast Dismantle on one of them, and then cast Radiate to get ω copies of Dismantle. Use Gleemax's power (this is the dubious bit) to make all the Dismantles target your indestructible artifact. Each resolving Dismantle will double the number of counters on your artifact, which means it ends up with 2^ω, i.e. ω₁.

If you're OK with playing an infinite number of activated abilities, then rather than using Gleemax, you could also do this trick with an infinite number of token copies of something like Torchling.

Isn't there some red enchantment which does twice as much damage?
There is. In fact, there are quite a few: Gratuitous Violence, Furnace of Rath, Anthem of Rakdos, Desperate Gambit (potentially useful with Radiate and an infinite number of creature tokens? Infinite coin-flips!), Impulsive Maneuvers(sic), and Overblaze (handy for doing ω₁ damage to someone).

Searching yields also Doubling Cube and Solarion, which each have potential. See also the Reflection cycle from Shadowmoor, specifically Boon Reflection, Mana Reflection and Thought Reflection.

[gareth-rees.livejournal.com]

2^ω, i.e. ω₁

I think you're confusing ordinal exponentiation with cardinal exponentiation here. Under the usual definition of ordinal exponentiation, 2^ω = ω, not ω₁.

(Even using cardinal exponentiation, 2^|ω| = 2^ℵ₀ = c, the cardinal of the continuum, not ℵ₁, unless you are assuming the continuum hypothesis.)

[alextfish.livejournal.com]

Bleh. It's too long since I did infinity theory. :) Yes, OK, I
was using ω where I meant ℵ₀, because I found an ω to copy-paste before I found a ℵ :)

I do tend to assume the continuum hypothesis. It simplifies things, because nobody's given any cardinal between ℵ₀ and c that's any use for ordinary discussions, and talking about ℵ₀ and ℵ₁ is more intuitive than talking about ℵ₀ and c.

[gareth-rees.livejournal.com]

Nobody's given any cardinal between ℵ₀ and c

Well of course they can't! (If they could then the continuum hypothesis would be false.)

ℵ₁ is the number of countable ordinals; that is, the number of well-orderings of the natural numbers. That doesn't seem particularly exotic.

[alextfish.livejournal.com]

Ah, and I've just remembered that he was trying to do this without ever paying ∞ mana into an X cost or playing an infinite number of abilities. He did it by tapping Mox Lotus with Transcendence and Soul Echo in play, letting the "gain ∞ life" ability go on the stack, then destroying Transcendence; then playing Storm Herd to make ∞ Pegasi.

Also, the Juniper Order Ranger is unnecessary.

[cartesiandaemon.livejournal.com]

Yeah, that makes sense. I think we'd end up allowing both of those, but doing it without makes it a more incontrovertible demonstration that it has to be possible.

[cartesiandaemon.livejournal.com]

Gleemax, ROFL. You might as well have gleemax in play if you've got infinite mana floating about anyway, but you're right that seems dubious: surely gleemax means you choose tagrets when targets are chosen? Admittedly, the only ruling I see isn't definitive, but it says "When does the controller of Gleemax choose the target? At the time the target would normally be chosen."

But I hope we can work round that.

[gareth-rees.livejournal.com]

I think the infinite number of doubling seasons produces an uncountable number of new counters.

I think not, since I can label each counter with a different integer. I label the first counter with 2. Then I label the next two counters 3 and 9 = 3². Then the next four counters 5, 25 = 5², 125 = 5³ and 625 = 5⁴. At doubling step k I label the 2^k new counters with the first 2^k powers of the (k+1)th prime number. The use of prime powers guarantees that the labels are distinct, and every counter gets labelled.

So the counters are countable. (Pun intended.)

[cartesiandaemon.livejournal.com]

Oh yes. Thank you for bringing some rigour, I was a bit asleep when I was thinking about this. And fuzzy on infinities, though I shouldn't be.

[cartesiandaemon.livejournal.com]

Indeed, I see I provided a helpful numbering system which numbered each of my counters with a rational real number. Doh! :) ETA: (Not even all rational numbers.)

[cartesiandaemon.livejournal.com]

BTW, thanks for reading through that :)

OK, I think I see the problem. I assumed counter number #0.01001000100001... was eventually created, whereas if you apply aleph-0 doubling effects, it is created at no finite time, so if you assume the set of tokens at each finite time to be supersets of each other, and the final result to be the union, then it does not contain #0.010010001...

On the other hand, it's a stack, so the effects should be applied in reverse order. If you could have effects at times (...,n,...,3,2,1). Thus at time n, you assume you have counters number 0.(0..0ⁿ)x, with n leading zeros, and x a real number. And then create counters 0.(0..0^n-1)1x at time n. That was all real numbered counters do get created (is that right)?

However, I don't know if the reverse order thing is well-defined. Indeed, it's not defined at all yet, I don't know if the result is always obvious, or even calculatable, or if it may be nonsense, or depend on the continuum hypothesis or something.

[alextfish.livejournal.com]

Doubling Season doesn't actually use the stack. It just replaces the action "put a token into play" with "put two tokens into play".

I don't think your proposed proof by induction works. I think the inductive step is fine, but your initial step is somewhat missing. (And in fact undefined, since you're trying to induce downwards from infinity to zero, which doesn't work too well.)

I've got myself a little confused about this, since I can't see why the Dismantle example above can produce 2^ℵ₀ if the set of ℵ₀ Doubling Seasons doesn't. You do seem to have provided an injection from your infinitely-Doubled tokens into the rationals, but equally it seems to me that "take 1 and double it ℵ₀ times" is what you're doing, and surely that must produce 2^ℵ₀ tokens?

[gareth-rees.livejournal.com]

take 1 and double it ℵ₀ times

Incidentally, I'm not at all sure what you mean by this. You're switching between ordinals and cardinals rather carelessly and this may be what's confusing you.

In particular, taking a number and doubling it repeatedly is a series of operations that have to happen in some order. So you must count the doublings with an ordinal, not a cardinal.

The order matters because you might get a different result if you do the doublings in a different order. For example, even though |ω| = |ω+1| = ℵ₀, we have 2^ω = ω, but 2^ω+1 = 2ω.