Solution for Exercise 0.3.25 in Jirka_RA

[ceciliachan1979]

This pull request provides the solution to exercise 0.3.25 in Basic Analysis I: Introduction to Real Analysis, Volumes I by Jiří Lebl.

https://www.jirka.org/ra/html/sec_basicset.html

[gapry]

The requirement is to show a bijection between S \ A and S.

If we assume that the domain is N and the range is also N, there seems to be no need to prove it, since it is guaranteed to be bijective if the domain and the range are the same set.

[ceciliachan1979]

The fact that $S$ contains a countably infinite subset implies there is a subset, let's call it $S_N$, apparently $S_N \subseteq S$ and there exists a bijective mapping from $\mathbb{N} \to S_N$, by the definition of countably infinite. in particular, the range is not $\mathbb{N}$, the range is $S_N$.

[Kelvinyu1117]

A should be $S_E$?

[ceciliachan1979]

No, I intend $A = S_O$. The idea is that:

The image of even natural numbers can map bijectively to Image of all natural numbers by
Step 1. Use $f^{-1}$ to get back to the even natural numbers
Step 2. Use $/2$ to get to all natural numbers, and finally
Step 3. Use $f$ to get to $S_N$.

So $S_E$ map bijectively to $S_N$.

$R$ obviously map to itself trivially using the identity mapping.

Together I achieved a bijective mapping that map $(S_E \cup R) = (S \setminus S_O) \to S$.

[Kelvinyu1117]

I'm not quite understand where $2f^{-1}(x)$ comes from?

[ceciliachan1979]

To prove the mapping is surjective, for every possible $x \in S$, I need to find an input (let's say $y$ )to $h$ so that $h(y) = x$, because I designed $h$, I know $y=f(2f^{-1}(x))$ will work out, so I just show that to you it does.

[E7-87-83]

for some x, y in S_E

[ceciliachan1979]

但係 for some express 唔到任何一對都 ok ⋯ ？

我諗咁寫會好啲

For any $x, y \in S_E$ such that $h(x) = h(y)$ , we have: