pi

For any finite sequence of numbers, that sequence can be found in the decimal extension of pi. True or false.

Ultimately, I don't think there is a true way to prove this.  There's no NP-Complete algorithm that will safely identify if any potential member x is in set Pi, when Pi itself never seems to end or wrap-around itself.  Other irrational numbers may offer clues to the secret of Pi inclusiveness.

A number 0.99... is irrational, but will only ever contains all sequences of the number 9 repeated x times.  In fact, it contains ALL combinations of any number of 9s repeating (an uncountable number of times).

An number like 0.7272... is irrational, but only contains sequences of 27, 72, and all numbers that are 27* or 72*.  It will never contain the sequence 277, despite that also being made of the digits 2 and 7.  A sequence that can be produced from the members of a member of a set does not have to be in the set.

The former indicates that an irrational number can contain every combination of the members of its set; and, the latter indicates that it will not necessarily contain a combination of the members of any single member in that set.  0.99... suggests that all the members of a set can be combined and represented in the irrational number, and we are not restricted by larger members such as in 0.7272... which creates roadblocks.  For the moment, all we can say is that Pi must necessarily contain as its alphabet all the natural numbers.  The "non-repeating" part of the definition is deceptive to us.