# Do I need a function to show a bijection between intervals?

1. Sep 27, 2012

### SMA_01

1. The problem statement, all variables and given/known data

I need to prove that [0,1) and (0,1] have the same cardinality. My question is, do I have to define a function from [0,1) → (0,1] in order to show a correspondence or is there another method?

Thanks.

2. Sep 28, 2012

### jbunniii

There's actually a very simple bijection in this case. (Hint: most of the points can simply be mapped to themselves.)

More interesting is to find one between (0,1) and [0,1].

3. Sep 28, 2012

### SMA_01

Would f(x)=1/1+x work? How would I find a bijection for the intervals you mentioned?

4. Sep 28, 2012

### Ray Vickson

No, f(x) = 1/1+x = 1 + x would not work, but f(x) = 1/(1+x) would be OK. Use brackets!

RGV

5. Sep 28, 2012

### SMA_01

Oh yeah, that's what I meant

In regards to the interval (whether open or closed), how can I be sure that the function will works as a bijection?

6. Sep 28, 2012

### jbunniii

If you can find an inverse function, then it's a bijection. To do that, put y = 1 / (1 + x) and try solving for x in terms of y.

There's also a much simpler solution if you consider the hint I mentioned above.

 Wait a minute. Who says 1 / (1 + x) will work? It maps [0, 1) to (1/2, 1], not (0, 1]. You'll have to modify it slightly to get (0,1] as the image.

7. Sep 28, 2012

### SMA_01

jbunniii- How did you know that 1/(1+x) maps [0,1) to (1/2,1] and not (0,1]?

And is the function you were hinting about f(x)=x+1?

8. Sep 28, 2012

### jbunniii

If 0 <= x < 1, then 1 <= 1+x < 2, so 1/2 < 1/(1+x) <= 1.

If 0 <= x < 1, then 1 <= x+1 < 2, so that won't work. If you put change it slightly by putting a negative sign in the right place, it will work.

Last edited: Sep 28, 2012
9. Sep 28, 2012

### SMA_01

I understand now, and the function you were referring to is f(x)=-x+1.

Thanks :)

10. Sep 28, 2012

### jbunniii

Yep. Here's an even easier one, the one I was thinking of initially:

$$f(x) = \begin{cases} 1 & \text{if }x = 0 \\ x & \text{if }x \neq 0 \end{cases}$$
By the way, if you want a challenge, I highly recommend trying to come up with a bijection between (0,1) and [0,1]. Hint: you know it can't be a continuous function because such a function would have to preserve open and closed sets.

11. Sep 28, 2012

### SMA_01

Cool, I didn't think of that. I am actually working on that now. I have to show that [0,1] has the same cardinality as (0,1).
I have a mapping:
0 to 1/2
1 to 1/3
1/2 to 1/4
etc.
So f(x)=1/2, if x=0
f(x)=1/(x+2), if x=1/n for positive integers n >= 1
and f(x)=x for all other x.
Is it correct syntax if I write it as a piece wise function?

f(x)={1/(x+2), if x is in N$\cup${0}
x, for all other x
I'm not sure how to word that last part (f(x)=x).

Edit:
This seems cleaner:

f(x)={1/(x+2), for x in NU{0}
x, for x in ℝ\ NU{0}

Where N denotes the set of all positive integers.
That seems cleaner.

Last edited: Sep 28, 2012
12. Sep 28, 2012

### jbunniii

Careful with your formula. For example, if x = 1/3, your formula gives f(1/3) = 1/(1/3 + 2) = 1/(7/3) = 3/7, which is not what you want. Instead, you could write $f(1/n) = 1/(n+2)$ for all $n \in \mathbb{N}$, $f(0) = 1/2$, and $f(x) = x$ for all other $x$.

No, your "if x is in $\mathbb{N} \cup \{0\}$" is wrong. That set is $\{0, 1, 2, 3, \ldots)$. But most of these points are outside your domain, which is $[0,1]$

Similar problem here.

13. Sep 28, 2012

### SMA_01

I see my mistakes. If I use f(1/n) and f(x), then I wouldn't write it in the format used for piecewise functions?

Also, to show the bijection, would I need to go through proving surjectivity/injectivity?

14. Sep 28, 2012

### jbunniii

I think you will still need a piecewise definition. It's hard to imagine a single formula working for all x. If you want to write it all in terms of f(x), you could put something like
$$f(x) = \begin{cases} \frac{1}{(1/x) + 2} & \text{if }x = 1/n \text{ for some }n \in \mathbb{N} \\ 1/2 & \text{if }x = 0 \\ x & \text{otherwise} \end{cases}$$
Yes, you should prove this. You could do this by showing that $f$ has an inverse.

15. Sep 29, 2012

### SMA_01

Okay, I did it. I let g be the inverse of f, defined by:

g: (0,1)→[0,1]

g(x)= { 0, if x=1/2
$\frac{1}{(1/x)-2}$ , if x=1/n for n$\in$ N
x, for all other x

Then I would prove that it's an inverse by composing g$\circ$f and f$\circ$g

Does that look fine?

Thanks.

16. Sep 29, 2012

### jbunniii

It looks almost right, except the middle case is only valid for n >= 3. The first case handles n = 2. And n = 1 is outside the function's domain.

17. Sep 29, 2012

### SMA_01

Oh yeah, I had that written down but completely missed it. Thank you for all your help