TITLE: Are there fewer reals on $(0, 1)$ than on $(1,\infty)$?
QUESTION [6 upvotes]: I know that the cardinality of the sets of real numbers $(0, 1)$ and $(1, \infty)$ are equal.
So what is the fallacy in this argument?

For every real on $(0, 1)$, we can add any integer $n$ to it and get a number on $(1, \infty)$, with the number from the first set as the fractional part of our new number. However, this can be applied to every real on $(0, 1)$, with every integer greater than one... seeming to suggest a larger cardinality of the second set than the first.

REPLY [3 votes]: The essential mistake here is the assumption that if you can establish a One-To-Many mapping from A to B, then you have proven that Cardinality(A) < Cardinality(B), or just in general, that they are not equal.  While this happens to be true for finite sets, this most definitely is not true for sets of transfinite cardinality.
Rather, you must rely on the definition of Equality for set-cardinality, which is that if you can establish any 1-to-1 mapping from A onto B then Cardinality(A) = Cardinality(B).  Therefore, to establish non-equal cardinality, you must show that there is no 1-to-1 onto mapping possible between them (which is what the Diagonal Proof does for Natural numbers Vs. Real numbers).