How does one reconcile the following (seemingly) contradiction in using "number of elements" argument?
In the "range" [0,1] in R there are more points than in N, to be shown as "take the inverse of every element of N and you end up with holes in [0,1]" This seems a valid argument.
Similarly it seems that [0,2] has twice the number of points as [0,1] Maybe it is invalid to claim this, but then why would the above be valid?
Is it fair to say that $f(x)= y = 2x$ with domain [0,1] and range [0,2] is bijective? no holes on the x-axis and no holes on the y-axis, mapped one-to-one but y-range still looks like to have twice the (infinite) number of elements. Does cardinality come to the rescue here, but then what about the one-to-one part of used for explaining bijectiveness?
I find this confusing, but I am sure many around here can clear this up, for this simple example. Thanks in advance.
While it is true that the cardinality of $\mathbb{N}$ is smaller than that of $[0,1]$ your argument is not a proof for. The former set is (infinite) countable while the latter has the cardinality of the continuum which is greater. This can be shown in a not too complicated way, yet there is some work to be done, going quite a bit beyond your argument.
The map $x \mapsto 2x$ is indeed a bijection from $[0,1]$ to $[0,2]$. The two sets have the same cardinality, the cardinality of the continuum.