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

  • #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 ?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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 ?
 
  • #4
PeroK
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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.
 
  • #5
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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.
 

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