Hi,

the point regarding the finite universe is well made indeed! I assume that the number of elements will follow this sequence:

NestList[2^# &, 1, 5]

A function which Mathematica evaluates easily. As you say the last number is quite large. If we display the logarithm to base 10 of these numbers we get:

Log[10, #] & /@ NestList[2^# &, 1, 5] // 
(*{0., 0.30103, 0.60206, 1.20412, 4.81648, 19728.3}*)

The last element has 19729 digits. Given that WolframAlpha says

== Number of atoms in the Universe

that there are only about $10^{80}$ atoms in the universe it might be difficult with the little matter in our Universe to actually print out these about $10^{19728}$ elements. Also, even if we wanted to display them, say each for a millisecond, it might -depending on your favourite model of the universe - take quite a bit longer than the Universe, as we know it, will exist. Well, there is a theory that says that if we wait a bit longer, $10^{10^56}}$ years or so, there might be a new big bang and we can start all over again. Even though the authors themselves seem to doubt that number.

Best wishes,

Marco

PS: If I ask Mathematica to calculate:

Log[10, #] & /@ NestList[2^# &, 1, 6] 

it bails out on me.

I think the problem is that the empty set has those pesky borders, one on each side. So we have to deal with power sets of braces, and that leads to a brace of power sets. How many to a brace? I always wonder about such things (not that I expect I'll ever be hunting pigeons in the countryside of whatever parts of Olde England have people hunting them by the brace). In this case, I guess because we have two borders that preceed and follow one power set, it follows that the result would be a set of two to the power of its predecessor. So that gets big fast.

Possibly I should have waited until March 32 (aka "two-to-the-fifth day" since March is the only month with that many days) to post..

Hi Daniel,

I think that the $6.655582364766226\cdot10^{19728}$ was for the fifth iteration as I said in my post. I cannot get the corresponding number for the 6th iteration....

Cheers,

M.