# Is tetration and quintation etc. of infinite cardinals well defined?

I would like to know whether tetration and quintation functions are well defined for infinite cardinals, thus, for example, $$\aleph_0 \text{ [tet] } \aleph_0 = \aleph_0 ^ {\aleph_0 ^ {\aleph_0^{\dots}}}$$ taken $\aleph_0$ times and compounded from top to bottom, to produce a higher cardinality power set upon every recursive iteration.

In an analogous manner, $$\aleph_0 \text{ [quint] }\aleph_0 = \aleph_0 \text{ [tet] } (\aleph_0 \text{ [tet] } (\aleph_0 \text{ [tet] } \dots))$$ taken recursively $\aleph_0$ times, again compounded from top to bottom.

Since as one moves to ever higher compouding operations like tetration, quintation, sextation etc. the series of new cardinals so producded would cleary increase MUCH faster than under an operation as "slow" as "mere" exponentially recursive power set taking, do we perhaps get the production of new huge cardinals that are of a different order and character than those generated by applying infinite power set operartions? Would all of these much larger cardinals, including even $\aleph_0\, [\aleph_0\text{ compounding operation}] \,\aleph_0$, itself compounded $\aleph_0$ times etc. still be smaller than other better known large cardinals defined in some other manner?

I hope these questions are fairly easy to comprehend. They hardly go much beyond elementary set theory. Thanks for reading and let me know if you can answer any of them!

• Please use 'MathJax' for equations – Eddy Nov 21 '17 at 21:37
• As far as I know there is no natural way to define an infinite tower of cardinal exponentiation. You could define it as a limit, but that is inconsistent with how infinite products are defined as distinct from a limit of finite products. – Eric Wofsey Nov 21 '17 at 21:38
• I once asked a similar question about ordinal arithmetic; the Op might find it interesting. math.stackexchange.com/questions/1359801/… – Sheldon L Nov 22 '17 at 1:33
• Thanks for the reference, Sheldon! I did find it very interesting. It reminded me of how I once mentioned the ordinal "epsilon omega" to one of my math colleagues only to hear him stammer back emotionally; "What?! Not that damn thing again!! I was up for three days straight trying to prove to myself that it must still be countable!!" – Wd Fusroy Nov 25 '17 at 1:26

## 1 Answer

dot dot dots are problematic in expressions because we require all expressions to be finite in length. A reasonable way to interpret the first part of your question is $$\aleph_0 \text{ [tet] } \aleph_0 = \sup\{\aleph_0, \aleph_0 ^ {\aleph_0},\aleph_0 ^ {\aleph_0 ^ {\aleph_0}},\dots\}$$ This is less than or equal to the first strong inaccessible (assuming it exists). A standard result is that if the exponent is infinite we can replace any base less than or equal to the cofinality of the exponent with $2$ so now we have $$\aleph_0 \text{ [tet] } \aleph_0 = \sup\{\aleph_0, 2 ^ {\aleph_0},2 ^ {2 ^ {\aleph_0}},\dots\}$$ If $\kappa$ is strongly inaccessible, we have $2^\alpha \lt \kappa$ whenever $\alpha \lt \kappa$ so the $\sup$ is at most $\kappa$. It must be lower because this is an increasing chain of length $\aleph_0$ so the cofiniality of the $\sup$ is $\aleph_0$ and the $\sup$ is not inaccessible.

• Hmm, why can't we take $\cup_n A_n$ with $A_n = 2^{A_{n-1}}$, $A_1 = \mathbb{N}$ within ZFC? Doesn't this show the $\sup$ is an accessible cardinal? – Jair Taylor Nov 22 '17 at 6:12
• @JairTaylor: You are right. It has to have cofinality $\aleph_0$. – Ross Millikan Nov 22 '17 at 6:23
• That makes sense, thanks! I was not previously familiar with cofinality. – Jair Taylor Nov 22 '17 at 6:29
• @JairTaylor . You can't "reach" an inaccessible cardinal by recursion from "below" it. For one thing, if ZF (or ZFC) is consistent then it cannot prove that an inaccessible exists. – DanielWainfleet Nov 23 '17 at 2:16
• Thanks so much to all of you who answered my questions!! I find all of the answers helpfulful. I will now go into the books and onto the web to try to find out just what has been written about this topic. I am actually rather surprised that it isn't treated even in most basic set theory books and courses, since as soon as one learns that exponentiation via the power set operation creates ever larger cardinals -- which adding and multiplying do not -- it's only natural to wonder if even "faster" compoundings do something similar aa well. – Wd Fusroy Nov 23 '17 at 15:07