- #1

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## Homework Statement:

- Prove that ## (0,1] \thicksim (0,1) ##

## Relevant Equations:

- Definition of a bijection

So, I need to prove that ## (0,1] \thicksim (0,1) ##. Which means that I need to come up with some bijection from ##(0,1]## to ##(0,1)##. Now here is the outline of my function. I am going to use identity function for all irrationals. So, any irrational number in ##(0,1]## will be mapped to the same number in ##(0,1)##. Now, I am going to divide rationals of ##(0,1]## as following. All rationals will be expressed in their lowest form. Now, with this understanding, first sub set is all rationals with numerator 1, second sub set is all rationals with numerator 2 and so on. Also, these sub sets will be disjoint. ##2/4## will not be included in the sub-set where numerator is ##2##, since this is ##1/2## and is already included in the first sub set. So, I have

$$ A_1 = \left\{ 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4}, \cdots \right\} $$

$$A_2 = \left\{ \frac{2}{3}, \frac{2}{5}, \frac{2}{7}\cdots \right\} $$

$$ A_3 = \left\{ \frac{3}{4}, \frac{3}{5}, \frac{3}{7}\cdots \right\} $$

I also define the following subsets of ##(0,1)##

$$ B_1 = \left\{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\cdots \right\} $$

$$B_2 = \left\{ \frac{2}{3}, \frac{2}{5}, \frac{2}{7}\cdots \right\} $$

$$ B_3 = \left\{ \frac{3}{4}, \frac{3}{5}, \frac{3}{7}\cdots \right\} $$

Now, I am going to map ##A_i## to ##B_i## for all ##i \geqslant 2##. So, this is going to be an identity function. And for ##A_1## and ##B_1##, the mapping will be as follows

$$ 1 \longrightarrow \frac{1}{2} $$

$$ \frac{1}{2} \longrightarrow \frac{1}{3} $$

$$ \frac{1}{3} \longrightarrow \frac{1}{4} $$

$$\cdots$$

So, with this kind of mapping, the mapping is a one-to-one and onto function from ##(0,1]## to ##(0,1)##. And that proves that ## (0,1] \thicksim (0,1) ##. Do you think this is a valid proof ?

$$ A_1 = \left\{ 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4}, \cdots \right\} $$

$$A_2 = \left\{ \frac{2}{3}, \frac{2}{5}, \frac{2}{7}\cdots \right\} $$

$$ A_3 = \left\{ \frac{3}{4}, \frac{3}{5}, \frac{3}{7}\cdots \right\} $$

I also define the following subsets of ##(0,1)##

$$ B_1 = \left\{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\cdots \right\} $$

$$B_2 = \left\{ \frac{2}{3}, \frac{2}{5}, \frac{2}{7}\cdots \right\} $$

$$ B_3 = \left\{ \frac{3}{4}, \frac{3}{5}, \frac{3}{7}\cdots \right\} $$

Now, I am going to map ##A_i## to ##B_i## for all ##i \geqslant 2##. So, this is going to be an identity function. And for ##A_1## and ##B_1##, the mapping will be as follows

$$ 1 \longrightarrow \frac{1}{2} $$

$$ \frac{1}{2} \longrightarrow \frac{1}{3} $$

$$ \frac{1}{3} \longrightarrow \frac{1}{4} $$

$$\cdots$$

So, with this kind of mapping, the mapping is a one-to-one and onto function from ##(0,1]## to ##(0,1)##. And that proves that ## (0,1] \thicksim (0,1) ##. Do you think this is a valid proof ?