Proving Injectivity of Shroeder-Berstein Theorem for R to R^2 x R^2

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Homework Statement


Prove that ##\mathbb{R}^2 \sim \mathbb{R}##

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The Attempt at a Solution


We will us the Shroeder-Berstein Theorem, and start with the simpler problem of showing that ##(0,1) \sim (0,1) \times (0,1)##. Define ##f: (0,1) \rightarrow (0,1) \times (0,1)## where ##f(x) = (x, \frac{1}{2})##. This is obviously an injection. Now define ##g: (0,1) \times (0,1) \rightarrow (0,1)## where if we suppose that every real number in this interval has a non-terminating decimal representation, ##g(0.x_1x_2x_3..., 0.y_1y_2y_3...) = 0.x_1y_1x_2y_2x_3y_3...##.

I just want to make sure that I am on the right track so far. How would I show that ##g## is an injection? Also, I know that in these decimal representations, we sometimes come into problems when the expansion ends in an infinite string of 9s. Do I run into that problem here, or am I good?
 
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By ##\mathbb R\sim\mathbb R^2## do you mean 'is bijective to'? I haven't seen that symbol used to mean that before, but from the context it seems that's what's meant.

It can (and should) be proven that any number that has a dec expansion that ends in an infinite string of nines also has a terminating dec expansion (or, put differently, an expansion that ends in an infinite string of zeros).

Having proven that, we can define ##g## such that for both inputs, where the input has an infinite-nines expansion, the infinite-zeros expansion is used instead.

The standard approach to proving injectivity is to assume that ##g(a_{11},a_{12})=g(a_{21},a_{22})## where all ##a_{jk}\in\mathbb R##, and then prove that ##a_{1i}=a_{2i}## for ##i\in\{1,2\}##.

If you write ##a_{jk}=\sum_{r=1}^\infty a_{jkr} 10^{-r}## where ##a_{jkr}\in\{0,1,...,9\}## then it becomes a problem in infinite series.
 
andrewkirk said:
By ##\mathbb R\sim\mathbb R^2## do you mean 'is bijective to'? I haven't seen that symbol used to mean that before, but from the context it seems that's what's meant.

It can (and should) be proven that any number that has a dec expansion that ends in an infinite string of nines also has a terminating dec expansion (or, put differently, an expansion that ends in an infinite string of zeros).

Having proven that, we can define ##g## such that for both inputs, where the input has an infinite-nines expansion, the infinite-zeros expansion is used instead.

The standard approach to proving injectivity is to assume that ##g(a_{11},a_{12})=g(a_{21},a_{22})## where all ##a_{jk}\in\mathbb R##, and then prove that ##a_{1i}=a_{2i}## for ##i\in\{1,2\}##.

If you write ##a_{jk}=\sum_{r=1}^\infty a_{jkr} 10^{-r}## where ##a_{jkr}\in\{0,1,...,9\}## then it becomes a problem in infinite series.
Could I argue that two integers are equal iff their decimal representations are equal, to conclude that ##g(a,b) = g(c,d) \Longrightarrow 0.a_1b_1a_2b_2a_3b_3... = 0.c_1d_1c_2d_2c_3d_3... \Longrightarrow (a_n = c_n) \wedge (b_n = d_n) \Longrightarrow (a=c) \wedge (b=d)##?
 
Mr Davis 97 said:
Could I argue that two integers are equal iff their decimal representations are equal
No. It's easy to prove the IF direction, but the ONLY IF direction is not valid, as the example ##1=0.\dot 9## demonstrates. The argument needs more careful tailoring. Also note that the term 'their decimal expansions' is inappropriate as it implies - incorrectly - that every number has a unique decimal expansion.
 
andrewkirk said:
No. It's easy to prove the IF direction, but the ONLY IF direction is not valid, as the example ##1=0.\dot 9## demonstrates. The argument needs more careful tailoring. Also note that the term 'their decimal expansions' is inappropriate as it implies - incorrectly - that every number has a unique decimal expansion.
Well if decimal representations aren't unique, how could I ever show that the function is injective?
 
andrewkirk said:
No. It's easy to prove the IF direction, but the ONLY IF direction is not valid, as the example ##1=0.\dot 9## demonstrates. The argument needs more careful tailoring. Also note that the term 'their decimal expansions' is inappropriate as it implies - incorrectly - that every number has a unique decimal expansion.
Also, I found this video:
And he uses the same method that I do.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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