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## 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.

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- Thread starter SMA_01
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- #1

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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

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

- #3

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

- #4

Ray Vickson

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Homework Helper

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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

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

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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.

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

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If 0 <= x < 1, then 1 <= 1+x < 2, so 1/2 < 1/(1+x) <= 1.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 <= 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.And is the function you were hinting about f(x)=x+1?

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

Thanks :)

Thanks :)

- #10

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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.

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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**N**U{0}

x, for x in ℝ\**N**U{0}

Where**N** denotes the set of all positive integers.

That seems cleaner.

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

x, for x in ℝ\

Where

That seems cleaner.

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- #12

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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].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?

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]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).

Similar problem here.Edit:

This seems cleaner:

f(x)={1/(x+2), for x inNU{0}

x, for x in ℝ\NU{0}

WhereNdenotes the set of all positive integers.

That seems cleaner.

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

- #14

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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 likeI 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?

[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]

Yes, you should prove this. You could do this by showing that [itex]f[/itex] has an inverse.Also, to show the bijection, would I need to go through proving surjectivity/injectivity?

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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]

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.

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- #17

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

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