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

[SMA_01]

Homework Statement

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.

 

[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].

 

[SMA_01]

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

 

[Ray Vickson]

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

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

RGV

 

[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?

 

[jbunniii]

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?

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.

[edit] 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.

 

[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?

 

[jbunniii]

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

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

And is the function you were hinting about f(x)=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.

 

[SMA_01]

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

Thanks :)

 

[jbunniii]

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

Thanks :)

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

[tex]f(x) = \begin{cases}
1 & \text{if }x = 0 \\
x & \text{if }x \neq 0
\end{cases}
[/tex]
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.

 

[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[itex]\cup[/itex]{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.

 

[jbunniii]

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?

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 [itex]f(1/n) = 1/(n+2)[/itex] for all [itex]n \in \mathbb{N}[/itex], [itex]f(0) = 1/2[/itex], and [itex]f(x) = x[/itex] for all other [itex]x[/itex].

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

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

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.

Similar problem here.

 

[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?

 

[jbunniii]

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?

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
[tex]
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}
[/tex]

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

Yes, you should prove this. You could do this by showing that [itex]f[/itex] has an inverse.

 

[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
[itex]\frac{1}{(1/x)-2}[/itex] , if x=1/n for n[itex]\in[/itex] N
x, for all other x

Then I would prove that it's an inverse by composing g[itex]\circ[/itex]f and f[itex]\circ[/itex]g

Does that look fine?

Thanks.

 

[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.

 

[SMA_01]

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