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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 $[a,b]$.

Sujaan Kunalan
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    For the first proof, $[0,1]$ is clearly a subset of $\mathbb{R}$. If it's uncountable, any set containing it must also be uncountable. For the second, there is a continuous bijection between $[0,1]$ and any closed interval $[a,b]$. – Roger Burt Nov 09 '14 at 01:47
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    @RogerBurt: I suggest you answer questions in the answer box, and not in the comments box. – Martin Argerami Nov 09 '14 at 01:54
  • @RogerBurt: Thanks. What is the difference between a bijection and a continuous bijection? – Sujaan Kunalan Nov 09 '14 at 01:57

2 Answers2

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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 $a<b$. In fact, the cardinality of $[0,1]$ is the same as that of $\mathbb R$.

Ittay Weiss
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$[0,1]$ is often just easier or cleaner to deal with.

With respect to your first example: the cardinality $\| [0, 1] \| = \| (0, 1) \| = \| \mathbb{R} \|$. (The second equality is straightforward, the first can be shown using Hilbert's Hotel, among many arguments.)

With respect to your second example, the obvious bijection $C[a,b] \rightarrow C[0,1]$ does everything we want.

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Simon S
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