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

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Homework Help Overview

The discussion revolves around proving the injectivity of the function as part of demonstrating the Schröder-Bernstein theorem, specifically in the context of showing that the set of real numbers, ##\mathbb{R}##, is bijective to ##\mathbb{R}^2##. The original poster attempts to establish this by first proving a simpler case involving the interval ##(0,1)## and its Cartesian product with itself.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of functions ##f## and ##g##, with the original poster seeking to confirm the injectivity of ##g##. There are questions about handling decimal representations, particularly regarding non-terminating expansions and their implications for injectivity. Some participants suggest clarifying the meaning of bijection and address potential issues with decimal representation uniqueness.

Discussion Status

The discussion is ongoing, with participants exploring various aspects of the problem. Some guidance has been offered regarding the treatment of decimal expansions, and there is an acknowledgment of the need for careful reasoning in proving injectivity. Multiple interpretations of the problem are being considered, particularly around the uniqueness of decimal representations.

Contextual Notes

There are constraints related to the assumptions about decimal expansions and the implications of non-unique representations, which are under discussion. The original poster is also navigating the requirements of the homework context while seeking clarification on specific mathematical concepts.

Mr Davis 97
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Homework Statement


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

Homework Equations

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.
 

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