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Response regarding inductive proofs and infinity
:::So pi is in pi for any large representation of pi, and that remains true as pi approaches infinity, right? This asymptotic approach of a subset of pi to pi itself is essentially the same as saying pi contains pi, right?--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 16:44, September 4, 2021 (EDT)
:::So pi is in pi for any large representation of pi, and that remains true as pi approaches infinity, right? This asymptotic approach of a subset of pi to pi itself is essentially the same as saying pi contains pi, right?--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 16:44, September 4, 2021 (EDT)
::::The book I'm reading (Charles Pinter's A Book of Abstract Algebra) seems to dodge the question - it has a section on inductive proofs, but it just starts with using integers and never mentions infinity. But the digits are countable, and go on for infinity, so they should be countably infinite - you should be able to match each digit to a digit in the countably infinite set of natural numbers, so that makes sense.
::::I was trying to avoid the term "subset" as by axiom, a [[set]] can't contain itself. But I wasn't thinking of this as a set anyway, as there are no duplicates in a set, so the set of all digits in pi would just be the finite set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 0\}</math> after removing duplicates.--[[User:AlChesapeake|AlChesapeake]] 20:22, September 4, 2021 (EDT)
The sentence in the article has been changed to
The sentence in the article has been changed to