Prove bijection from (0,1] to (0,1)

[issacnewton]

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 ?

 

[Mark44]

I don't see anything wrong with this. It's also possible to show that [0, 1] is isomorphic to [0, 1] X [0, 1], but the mapping, like yours, is of course not continuous.

 

[issacnewton]

Thanks Mark, of course this is not continuous. I think its not possible for the function to be continuous in this case. What do you think ?

 

[PeroK]

Thanks Mark, of course this is not continuous. I think its not possible for the function to be continuous in this case. What do you think ?

Wouldn't your proof be simpler if you just considered the set $A_1$? I.e. just the sequence $1, \frac 1 2 \frac 1 3 \dots$? Why involve all the other rationals?

In terms of discontinuity, you can construct a proof by considering the $a \in (0, 1)$ that maps to $1$; then use the intermediate value theorem to show that a continuous mapping from $(0,1)$ to $(0,1]$ cannot be one-to-one.

 

[issacnewton]

PeroK, yes, essentially, the proof boils down to the set $A_1$. But I could not think this initially. I had to divide rationals and irrationals first and then try to think of how the rational mapping is to be done.