6
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You might call me a number,

but that would be a blunder.

In class you may have been told otherwise,

but I tell you now that those were all lies.

I'm in the deck of the cards,

as big as them come.

You'll never count up to me,

I'm beyond such a sum.

You can only reach me with creativity and wit,

leaving the finite-minded in a rage and a fit!

My innards in size relate,

though you may debate,

what is larger than largest,

your intuitions are tarnished.

What am I?

Hint 1

enter image description here

Hint 2

Hint 1 is just a beginning, but you must go further, to keep your head spinning. Further and further and further and...

Hint 3

Let Burali-Forti be your guide, the path is long but not the ride.

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I believe, If I assume ZFC, or rather any other consistent set theory, the answer is :

Such a thing does not exist.

Reasons for the claim:

Let such a thing exist. Call it $C$.

You might call me a number, but that would be a blunder

It is obviously larger than any natural number as natural numbers are countable and thus bounded by the ordinal $\omega$, and $C>\omega$ because it is "larger than largest" (and so $C$ is greater than any known cardinality, which will form the basis of our contradiction).

I'm in the deck of the cards, as big as them come.

An allusion to the fact that $C$ is the cardinality of some set, and is very very large (please do not hate me)

You'll never count up to me, I'm beyond such a sum. You can only reach me with creativity and wit, leaving the finite-minded in a rage and a fit!

Since $C$ is larger than $\aleph_1$, we cannot count up to it, rather create hypothetical sets that have such a cardinality.

My innards in size relate, though you may debate

There is a relation between cardinalities smaller than $C$ and $C$. For example, $\aleph_1=2^{\aleph_0},\ \aleph_2=2^{\aleph_1}$ and so on and $\aleph_0<\aleph_1<\aleph_2$. It is debatable because in ZFC, one cannot prove that there is nothing between these cardinalities.

what is larger than largest, your intuitions are tarnished.

The final piece of the puzzle. $C$ is larger than any cardinality, which is analogous to the statement that $\omega$ is larger than any natural number. But can $C$ exist? Claim : No. Proof : If $C$ exists, then there is a set whose cardinality is $C$. Take the power set of this set. The power set has strictly larger cardinality than $C$ which contradicts the fact that $C$ is "larger than the largest". QED This actually throws our intuition off track, and we are left dazed.

Hint 2 tells us:

Keep going beyond what we can find. That's what we had done to create $C$ but unfortunately such a construction is not possible as we had seen.

Hint 3 is mathematical :

Burali-Forti is a very well known paradox that states you can't create the set of all ordinals, which seems unintuitive.

Eh?

A mathematical proof is, although not complicated, but involved, and here's how we can understand it. Let the ordinal associated with $C$ be $\Omega$. Now since ordinal numbers are well-ordered in a natural way, this ordering must have type $\Omega$. We know that the order type of all ordinal numbers less than some fixed ordinal number is the ordinal number itself. The, the order type of ordinal numbers $<\Omega$ is $\Omega$. But this means that $\Omega$, being the order type of a proper initial segment of the ordinal numbers, is strictly less than the order type of all the ordinal numbers, but the latter is itself $\Omega$, and we have reached a contradiction.

Okay...

And we know that such a set cannot be formed (at least in ZFC, but in some theories, it forms a class, but that's a different story), and hence has no cardinality. But this should be "larger than the largest". Hence proved.

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    $\begingroup$ You're getting warm! $\endgroup$ – Galen Apr 26 '20 at 14:42
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You seem to be talking about

$\aleph_1$ = Aleph one (cardinality of the real numbers)

You might call me a number,
but that would be a blunder.
In class you may have been told otherwise,
but I tell you now that those were all lies.

Aleph one is the cardinality of the real numbers although it might not be helpful to think of it as a number in the traditional sense. Often people think of it as equivalent to $\infty$ but while $\infty$ represents the limit of a divergent series, the infinite ordinals are a different concept.

I'm in the deck of the cards,
as big as them come.

I think "cards" is short for cardinalities and aleph one is in "the deck of cards". "As big as them come" might refer to the fact that these ordinals are of infinite size.

You'll never count up to me,
I'm beyond such a sum.

In ZF set theory, $\aleph_1$ is the first "uncountable" infinite ordinal.

You can only reach me with creativity and wit,
leaving the finite-minded in a rage and a fit!

I think this refers to Cantor's diagonal argument which is used to show that the cardinality of the real numbers is, in a concrete sense, greater than the cardinality of the natural numbers. It's hard to wrap your head around this when thinking from a finite perspective.

My innards in size relate,
though you may debate,

Open intervals of the real numbers are of the same cardinality as the real numbers themselves even though it might be debated that, as subsets, they are "smaller".

what is larger than largest,

The concept of infinite ordinals poses the question, "what is larger than infinity?"

your intuitions are tarnished.

Thinking that there are various "levels" of infinity might go against intuition.

Diagram

This shows symbols for the first 5 infinite ordinals with arrows indicating "smaller than" in cardinality ordering.

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    $\begingroup$ I believe this is the answer and what the OP has in mind, but something is bugging me from the very beginning. Larger than the largest? If we take it is $\aleph_1$, then it must be larger than the largest. But it is not as it is not even larger than $\aleph_2$. Contradiction. $\endgroup$ – Yuzuriha Inori Apr 26 '20 at 12:53
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    $\begingroup$ r13 Creuncf "va gur qrpx bs pneqf" ersref gb na 8 ghearq fvqrjnlf. $\endgroup$ – Thomas Markov Apr 26 '20 at 23:14
  • $\begingroup$ Tnyra'f uvag gung Uvag 1 vf whfg gur ortvaavat vaqvpngrf gung gurl znl jnag hf gb tb orlbaq nyy pneqvanyf. Pbhyq gur nafjre or gur havirefny frg (r.t. va Dhvar'f Arj Sbhaqngvbaf)? Jbhyq znxr gur Ohenyv-Sbegr ersrerapr znxr frafr. $\endgroup$ – Anon Apr 27 '20 at 4:07
  • $\begingroup$ rot13 Gung jbhyq znxr frafr bayl vs gur beqvanyf jrer jryy beqrerq va AS, juvpu vg vf abg. Gur "pneqvanyvgl" bs gur havirefny frg npghnyyl vf abg gnyxrq nobhg va AS, be gb or crqnagvp, qbrf abg rkvfg. $\endgroup$ – Yuzuriha Inori Apr 27 '20 at 5:53
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    $\begingroup$ @Anon Thank you for clarifying that. I used rot13 to read them. $\endgroup$ – Galen Apr 28 '20 at 0:14
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It’s a strange guess, but

Zero

I'm in the deck of the cards, as big as them come.

In the 10

You’ll never count up to me, I’m beyond such a sum.

The Natural “Counting” Numbers do not include zero

My innards in size relate,

Adding a zero, except trailing, represents a multiplication or division by 10

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I am simply going to go with

Infinity

"I'm in the deck of the cards, as big as them come."

This could be talking about the Joker, as if it is all some cosmic joke to keep trying to grasp such a magnitude.

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  • $\begingroup$ This doesn't fit well enough, but I really like where you went with that. +1 $\endgroup$ – Galen Apr 26 '20 at 17:29

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