Finite Sets of Infinite Things

[Treadstone]

had an idea in class the the other day...feel free to add on or modify for whatever, just a rabbit trail of thought....

Let A = { pi , 2^(1/2) }

A only contains 2 items and so Card(A) = 2 => there exists a bijective function from A to the natural numbers of order 2 => A is finite...probably didnt need to prove it but there it is.

Write the items in A as sets themselfs as such

pi = {3, 1, 4, 1, 5, 9, ... }
2^(1/2) = { 1, 4, 1, 4, 2, 1, ... }

So can A be rewritten as A = { {3 1 4 1 5 9...} , {1 4 1 4 2 1...} } ?

Would this imply that A has an infinite number of things dispite that it is a finite set? Or is it just my notation and how i'm defining things?

Any interesting ideas on where i could go from here?

[Aeschylus]

had an idea in class the the other day...feel free to add on or modify for whatever, just a rabbit trail of thought....

Let A = { pi , 2^(1/2) }

A only contains 2 items and so Card(A) = 2 => there exists a bijective function from A to the natural numbers of order 2 => A is finite...probably didnt need to prove it but there it is.

Write the items in A as sets themselfs as such

pi = {3, 1, 4, 1, 5, 9, ... }
2^(1/2) = { 1, 4, 1, 4, 2, 1, ... }

So can A be rewritten as A = { {3 1 4 1 5 9...} , {1 4 1 4 2 1...} } ?

Would this imply that A has an infinite number of things dispite that it is a finite set? Or is it just my notation and how i'm defining things?

Any interesting ideas on where i could go from here?

Well no, becasue it's the numbers themselves not the digits in their decimal representation (and it must be added thta choosing them to be represnted as decimals is completely arbitay) that are elements of the set.

Of course you could have a set A = {R ,R^2} this set has two elements yet it's elements are infinite sets. Of course there's nothing particualry remarkable about that as it is not the elements of the sets R and R^2 that belong to A, but the sets themselves.

[Treadstone]

ah i see....thanks for the clear explanation