markr wrote:raprap wrote:However, in my mind a logic exists. You could partially prove the theorem using two sets, one set (A) containing all numbers (natural or otherwise) from 5 to positive infinity (inclusive?) and the other (B) all numbers from 10 to positive infinity. The first set (A) then contains numbers that are not in the second one (B), whilst the intersection of the two sets (A intersection B=B) is the same as the second set (B). Consequently, there is no one to one relationship from the elements in A to B and the sets do not have the same cardinality (number of elements).
So set A is larger than B. QED(?)
Rap
No. There is a one-to-one mapping of set B to set A:
b -> b-5
The sets have the same cardinality. Your proof doesn't apply to infinite sets.
1) Raprap, an infinite subset has the same cardinality as its superset.
2) markr, raprap's logic is correct...in that looking for a one-to-one and onto mapping is the preferred way of considering the size of an infinite set...although their cardinality is the same, this does show that B is a subset of A.
3) "size" is ill-defined on infinite sets. Yes B is a subset, but since they have infinite size, you cant say it is equal to or greater than or less than the size of the other.
4) Infinity has a different meaning in the context of an infinite set than an infinite value, you cannot make it into an infinite set to prove distance between a number and an infinite value.
5) markr, yes there is a one-to-one mapping, but raprap meant to say one-to-one AND onto...and it's not onto.