I would like to know why the following are true. I believe it is some sort of relationship between the two spaces- an isomorphism of sorts?
There is a proof to show that the real numbers is uncountable. The proof uses $[0,1]$ instead of $\mathbb{R}$. Why can we work with $[0,1]$ instead of $\mathbb{R}$?
Another instance of this:
There is a proof of the Weierstrass Approximation Theorem, which can be proved by proving Bernstein's Theorem first. In the proof for Bernstein's Theorem, although we want to show that $B_n(f)$ converges to $f$ uniformly for all $f\in C[a,b]$, we merely suppose $f\in C[0,1]$ instead, where $(B_n(f))(x)=\sum_{k=0}^nf\left(\frac{k}{n}\right)\cdot
{n\choose k}x^k(1-x)^{n-k}$, for $x\in[0,1]$, and where $C[a,b]$ is the set of all continuous functions on the interval $[a,b]$.
Answer
The first result your mention is set-theoretic. It is easily verified that if $A\subseteq B$ and $A$ is uncountable, then $B$ is also uncountable. So, showing that $[0,1]$ is uncountable suffices to show the uncountability of the reals. For Cantor's diagonal argument it is easier to work with $[0,1]$, which is why we choose to do so. It should be noted the proof is typographical. It is about sequences of sequences of digits.
As for the second result, I'm not entirely sure what you are asking, but here there is no essential difference between any two closed intervals. Again, for Bernstein's proof (which is probabilistic) it is easier to normalize things and work with $[0,1]$. Using any linear function $[0,1]\to [a,b]$ then translates Weierstrass's approximation result to any closed interval. Note that then it also implies that the cardinality of $[0,1]$ is the same as that of $[a,b]$ for all $a
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