A mathematical zen koan

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Say you have a dot. That dot explodes into an infinite number of dots. Each of those dots explode into a new infinite number of dots. Each of those dots explode into a new infinite number of dots, ad infinitum.

How many dots do you have in total?
 
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mr3000 said:
Say you have a dot. That dot explodes into an infinite number of dots. Each of those dots explode into a new infinite number of dots. Each of those dots explode into a new infinite number of dots, ad infinitum.

How many dots do you have in total?
A shitload.
 
mr3000 said:
How many dots do you have in total?
= ∞
 
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My knowledge about infinities is infinitesimally small, but there is a chance answer depends on what kind of infinity you started with. Was it ##\aleph_0##? ##\aleph_1##? Something else?
 
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mr3000 said:
Say you have a dot. That dot explodes into an infinite number of dots. Each of those dots explode into a new infinite number of dots. Each of those dots explode into a new infinite number of dots, ad infinitum.

How many dots do you have in total?
Borek said:
My knowledge about infinities is infinitesimally small, but there is a chance answer depends on what kind of infinity you started with. Was it ##\aleph_0##? ##\aleph_1##? Something else?
That's right. I'm pretty sure the infinitizing operation doesn't change the index of the infinity. For natural numbers, have each infinity be assigned a different prime. Then each number in that infinity is a different power of that prime. And so forth.
 
mr3000 said:
Say you have a dot. That dot explodes into an infinite number of dots. Each of those dots explode into a new infinite number of dots. Each of those dots explode into a new infinite number of dots, ad infinitum.

How many dots do you have in total?
Mr3000 asks the same question ad infinitum.
When infinite time has passed, he shall have our answer.
 
PeroK said:
Mr3000 asks the same question ad infinitum.
When infinite time has passed, he shall have our answer.
...yet doesn't read a textbook on the cardinalities of infinite sets. The question is... is it because he Cantor won't?
 
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Ibix said:
...yet doesn't read a textbook on the cardinalities of infinite sets. The question is... is it because he Cantor won't?
A book that lies unopened conceals its wisdom.
 
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Hornbein said:
That's right. I'm pretty sure the infinitizing operation doesn't change the index of the infinity. For natural numbers, have each infinity be assigned a different prime. Then each number in that infinity is a different power of that prime. And so forth.
I disagree. Allowing the process to go on infinitely long can increase the size. Cantor's diagonal proof shows that the reals, represented by infinite sequences of digits, are an uncountable set.
 
The cardinality of the power set of a set is strictly greater than the cardinality of the set. The power set exists by axiom of ZFC, and the cardinality is greater by a generalization of Cantor's diagonalization argument (https://en.wikipedia.org/wiki/Cantor's_theorem).

So ##2^{\aleph_0}>\aleph_0##, ##2^{\aleph_1}>\aleph_1##,...

It's not clear to me if this is what the opening question is about though.
 
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mr3000 said:
Say you have a dot. That dot explodes into an infinite number of dots. Each of those dots explode into a new infinite number of dots. Each of those dots explode into a new infinite number of dots, ad infinitum.

How many dots do you have in total?

This question is a Zen Koan because the naive intuition says, “If each point explodes into an infinite iteration of infinitely many new points, surely the set gets bigger.” But infinite cardinal arithmetic says: “No ,not unless the new sets are strictly larger than the original.”

It's helpful to first review two historical examples: Georg Cantor showed the set of natural numbers are countable whereas the set of real numbers between zero and one is an uncountable set. These infinite sets have different sizes denoted by their cardinality which expresses how many members are in a set. The cardinality of the natural numbers is given by ##|\mathbb{N}|=\aleph_0## (alpha in Hebrew) and that of the reals between zero and one is given by ##|(0,1)|=2^{\aleph_0}##. What this means is that the "size" of the set of real numbers is equal to the size of the set of all subsets of natural numbers (the power set ##P(\mathbb{N}))##, that is, ##
|(0,1)|=|P(\mathbb{N})|=2^{\aleph_0}
##

Another fact about infinite sets is if you begin with an infinite set of cardinality ##K## and each point "explodes" into a set of the same cardinality ##K## then ##K\times K=K##.

And even repeating this infinitely many times, ##K\times K\times K\times\cdots=K##.

So the size or "cardinality" of the set under this operation does not change. This is because infinite cardinal arithmetic collapses multiplication. That is: ##K\cdot K=K## and ##K\cdot \aleph_0=K##.

However, if each point explodes into a set ##L>K##, then the first explosion produces a set of size ##L## or ##K\cdot L=L## and every subsequent explosion of size ##L## stays at size ##L## or ##L\cdot L=L## but only because the initial set was replaced by a set of larger cardinality.

The only way to guarantee replacing a set of cardinality ##K## with the smallest larger infinity is to replace a set with its power set ##2^{K}>K##. This is why the hierarchy ##K,2^K,2^{2^K},2^{2^{2^K}},\cdots## is the standard way to climb the ladder of larger and larger infinities.

We can then place limits on the size of the resulting set:

If we start with the smallest infinite cardinal, ##\aleph_0## and replace each element in this set by another ##\aleph_0## set, and do this infinitely often, then the size of the set remains at ##\aleph_0##.

However, if the first explosion begins with cardinality ##K## and then each point in this set is replaced by a set ##L>K##, then the resulting set, even if expanded infinitely often, remains at ##L##.

If we start with the smallest cardinal ##\aleph_0##, and then replace this set of points by the next cardinal ##2^{\aleph_0}## and replace that set by the next higher cardinal ##2^{2^{\aleph_0}}##, and do this infinitely often, the sequence ##\aleph_0,2^{\aleph_0},2^{2^{\aleph_0}},\cdots## has no largest element and no limit within the sequence, but its least upper bound (supremum) is the cardinal given by the symbol ##\beth_{\omega}## (beta in Hebrew).

Starting from ##\aleph_0##, the smallest possible cardinality of the exploded set is ##\aleph_0##. And if we allow replacement by successively larger cardinals in the power‑set tower, the supremum (least upper bound) of the possible sizes is defined as ##\beth_{\omega}##.
 
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