[jepler]

That may feel like an intuitive declaration, but that's not how it works in standard mathematics. In standard set theory, the cardinality of the reals is equal to the cardinality of any non-degenerate interval of reals. Wikipedia quotes this fact without proof (https://en.wikipedia.org/wiki/Cardinality_of_the_continuum#S...).

Here's one hand-wavy proof of why the cardinality of the real interval P=[0,1] is the same as the cardinality of the real interval Q=[0, infinity]: The function f(x) = 1/x-1 is a bijective function that maps the P interval onto the Q interval, which also proves the cardinality of the two sets is equal. (https://en.wikipedia.org/wiki/Bijection#Cardinality).

If you're not comfortable with 1/0 = infinity as a general matter, then simply replace the f(x) I gave with an explicit piecewise function f(x) = { 0 if x = ∞, else (1/x-1) } and the proof still works.