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## Main Question or Discussion Point

Hello, I'm working on some questions and I need some further explanation;

First I must

Question

So my first thought is the function [tex](\frac{1}{a},\frac{1}{b}) : a,b\in\mathbf{N}[/tex]. I think this function is into, but not all the members of the set S are mapped to, thus not being onto. Is this right?

The next question i had a problem with:

Heres my thought for (b), i define a set [tex]A_n , n\in\mathbf{N}[/tex] where n represents the number of decimal places. So n=1 would be one decimal place, n=2 would be two decimal places, and so on. With only one decimal place, [tex]A_1[/tex] contains every combination of (x,y) such that 0<x,y<1 for instance (.9,.3) or (.1,.5). [tex]A_2[/tex] contains every combination of (x,y) such that 0<x,y<1 to two decimal places, and so on.

So if i arrange all the [tex]A_n[/tex] like this:

[tex]A_1(1)[/tex] [tex]A_1(2)[/tex] [tex]A_1(3)[/tex] [tex]A_1(4)[/tex] ...

[tex]A_2(1)[/tex] [tex]A_2(2)[/tex] [tex]A_2(3)[/tex] ...

[tex]A_3(1)[/tex] [tex]A_3(2)[/tex] ...

[tex]A_4(1)[/tex] ...

...

and i arrange [tex]\mathbf{N}[/tex] like this:

1 3 6 10 ...

2 5 9 ...

4 8 ...

7 ...

...

Then S maps into [tex]\mathbf{N}[/tex] since i represented every decimal expansion in the array. Is this right? But I'm not sure how the reminating decimal affects this, since, for instance, .9 which belongs to [tex]A_1[/tex] is also .8999... .

Why would i apply something that follows to this question?

First I must

*Consider the open interval (0,1), and let S be the set of point in the open unit square; thats is, S={(x,y):0<x,y<1}.*Question

**says***(a)**Find a 1-1 function that maps (0,1) into, but not necessarily onto, S. (this is easy)*So my first thought is the function [tex](\frac{1}{a},\frac{1}{b}) : a,b\in\mathbf{N}[/tex]. I think this function is into, but not all the members of the set S are mapped to, thus not being onto. Is this right?

The next question i had a problem with:

*(b) Use the fact that every real number has a decimal expansion to produce a 1-1 function that maps S into (0,1). Discuss whether the formulated function is onto. (keep in mind that any terminating decimal expansion such as .235 represents the same real number .2349999... .)*Heres my thought for (b), i define a set [tex]A_n , n\in\mathbf{N}[/tex] where n represents the number of decimal places. So n=1 would be one decimal place, n=2 would be two decimal places, and so on. With only one decimal place, [tex]A_1[/tex] contains every combination of (x,y) such that 0<x,y<1 for instance (.9,.3) or (.1,.5). [tex]A_2[/tex] contains every combination of (x,y) such that 0<x,y<1 to two decimal places, and so on.

So if i arrange all the [tex]A_n[/tex] like this:

[tex]A_1(1)[/tex] [tex]A_1(2)[/tex] [tex]A_1(3)[/tex] [tex]A_1(4)[/tex] ...

[tex]A_2(1)[/tex] [tex]A_2(2)[/tex] [tex]A_2(3)[/tex] ...

[tex]A_3(1)[/tex] [tex]A_3(2)[/tex] ...

[tex]A_4(1)[/tex] ...

...

and i arrange [tex]\mathbf{N}[/tex] like this:

1 3 6 10 ...

2 5 9 ...

4 8 ...

7 ...

...

Then S maps into [tex]\mathbf{N}[/tex] since i represented every decimal expansion in the array. Is this right? But I'm not sure how the reminating decimal affects this, since, for instance, .9 which belongs to [tex]A_1[/tex] is also .8999... .

*The puzzling thing about this question is that at the end of the question it says this: The Schroder-Bernstein Theorem discussed in 1.4.13 to**can now be applied to conclude that (0,1)~S.***follow**Why would i apply something that follows to this question?

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