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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.
No, f(x) = 1/1+x = 1 + x would not work, but f(x) = 1/(1+x) would be OK. Use brackets!Would f(x)=1/1+x work? How would I find a bijection for the intervals you mentioned?
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.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 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?
Yep. Here's an even easier one, the one I was thinking of initially:I understand now, and the function you were referring to is f(x)=-x+1.
Thanks :)
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 in NU{0}
x, for x in ℝ\ NU{0}
Where N denotes the set of all positive integers.
That seems cleaner.
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?
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?