Real Analysis :Prove (0,1) is equivalent to [0,1]

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kathrynag

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Define f(0,1)-->R as follows
For n element of natural numbers, n>=2, f(1/n)=1/n-1 and for all other x elements of( 0,1), f(x)=x.

[0,1) is equivalent to (0,1]
I sort of have an idea about this one.
If A=[0,1), B=(0,1] and C=[0,1], then A=C-x1contained in C and B=C-x2
contained in C
A and B are equivalent because they both are C-some x. Is this somewhat
right? Cause this seems to make sense to me.

Now my problem is for (0,1) is equivalent to [0,1]. if I can use the C set idea I would really like that!
 
Since the one set is open and the other is closed, I'm not sure on what basis they would be "equivalent"...?

Eliz.
 
Ok I know [0,1) is equivalent to (0,1] by f(x)=1-x
Now I'm having trouble finding a similar function for (0,1) equivalent to [0,1]
 
kathrynag said:
It's not saying they are equal.
Click to expand...
They aren't the same set, so obviously they aren't "equal", as may be trivially shown by providing the two elements of the one set that do not exist in the other. The point of the previous reply, however, was not equality, but equivalence. You have one set that is open and another is closed. In other words, they differ in a fundamental property. How would they then be "equivalent"?

kathrynag said:
Just saying they are equivalent.
Click to expand...
What then is the definition of "equivalent" sets, word for word as given in your textbook?

Thank you! :D

Eliz.
 
stapel said:
kathrynag said:
It's not saying they are equal.
Click to expand...
They aren't the same set, so obviously they aren't "equal", as may be trivially shown by providing the two elements of the one set that do not exist in the other. The point of the previous reply, however, was not equality, but equivalence. You have one set that is open and another is closed. In other words, they differ in a fundamental property. How would they then be "equivalent"?

kathrynag said:
Just saying they are equivalent.
Click to expand...
What then is the definition of "equivalent" sets, word for word as given in your textbook?

Thank you! :D

Eliz.
Click to expand...

Poster means equipollent: the cardinalities of the two sets ARE the same. Since you have already proven (you say, and I believe you) that the cardinality of [0,1) is equivalent to (0,1] (by your concise bijection), there are still some pieces missing.

Therefore, if

(0,1) is equipollent to (0,1] ... call this (A) and
(0,1] is equipollent to [0,1) ... call this (B) and
[0,1) is equipollent to [0,1] ... call this (C)

then, doesn't it follow that (0,1) is equipollent to [0,1] (transitive property)? Now, you need to find bijection functions for (A) and (C). I believe yours for (B) is fine. And this is only one way to prove (0,1) is equipollent to [0,1], and it may involve more steps than some other ways. Does anyone else have any other pathways. I can't think of a single bijection function, which is what your original post seemed to be suggesting you were looking for. Sorry, but this may be a more thought-provoking exercise anyway.
 
No, equivalent means thast for 2 sets A and B, there is a 1-1 function from A onto B.
 
Ok, I'm gonna lay out what I've done thus far.


proving f is a 1-1 function from (0,1) into (0,1].
Let a=1/m b=1/n
f(a)=1/(m-1) f(b)=1/(n-1)
f(a)=f(b)
m=n
1/m=1/n
a=b

f(a)=a f(b)=b
f(a)=f(b)
a=b


proving f is a function from (0,1) onto (0,1]:
To prove f:A-->B is onto, let B be arbitrary and show there exists a in A such that f(a)=b. I need to show that im f=B.
y=1/(n-1)
n=1/(y-1)
y-1=1/n
f inverse=1/n+1
f inverse =(1+n)/n
f inverse of b=(1+b)/b=a
f(a)=1/[(1+b)/b-1]=1/1/b=b

y=x
x=y
f inverse=x
f inverse (b)=b=a
f(a)=a=b

Finding a 1-1 function from [0,1) onto [0,1]
[0, 1) and [0, 1]
given a bijection f:(0,1)-->(0,1] , define g:[0,1)-->[0,1] by g(0) =0, and g(x) = f(x) for x in (0,1)

Proving [0,1) is equivalent to (0,1]:
Use the function f(x)=1-x

Proving (0,1) is equivalent to [0,1]:
So then let f:(0,1)-->(0,1] and let g:(0,1)-->[0,1]
Let f(x)=g(x) for all x in (0,1)
Is this part correct? I know there needs to be some number to equal 0
right?
Let g(x)=f(x)=?
 
I'm quite surprised no one else has weighed in. I'm lost with all the parentheses or missing parentheses. Someone else is going to have to check your work at this point.

I just wanted to suggest you break it up, but now I think you are closer than you think. Actually, don't you have it by the transitive property?

I suggested the transitive property, because it's a way to break it up. If you can get simpler pieces, you can put them together like that. If thing A (namely, the cardinality of (0,1)) equals thing B (cardinality of (0,1]) AND thing B equals thing C (cardinality of [0,1)), then you know thing A equals thing C.

A single bijection seemed like it would be very difficult to find. And you've taken care of that, I think.

You have (0,1) eq. to (0,1]
and (0,1] eq. to [0,1)
and [0,1) eq. to [0,1]

Isn't that all you need? Are you sure your functions are correct?

Another idea: A function doesn't have to be a single expression. Have you thought about

f(x) = some expression in x, x < 1/2
= some other expression in x, x >= 1/2

?? You might need to do that.

(The reason we got hung up on the word "equivalent," I think, is that many authors use the word loosely. Recently, we have started using the word "equipollent" to mean the cardinality of two sets is equal. That's what you're referring to when you say there is a 1-to-1 function that maps set A onto set B. Many textbooks that I have seen still use "equivalent" for this purpose, but if you weren't talking about sets defined by open and closed intervals, people would have thought it was highly ambiguous. Please forgive our challenges to your vocabulary for textbooks that have not yet come up-to-date.)
 
Prove [0,1) is equivalent to (0,1)?
f(1/n)=1/(n-1) for n>=2
f(x)=x for all other elemnets of (0,1)
Ok I know I need to find a 1-1 function from [0,1) onto (0,1)
My problem isdoing this because 0 is not included in the range. Can someone please try to help? I need to present this tomorrow.
 
Why not do it directly and simply? \(\displaystyle f:\left[ {0,1} \right] \mapsto \left( {0,1} \right)\)
\(\displaystyle f(x) = \left\{ {\begin{array}{l,r} {\frac{1}{2},} & {x = 0} \\ {\frac{1}{3},} & {x = 1} \\ {\frac{1}{{n + 2}},} & {x = \frac{1}{n},\;n \geqslant 2} \\ {x,} & {else} \\ \end{array} } \right.\)

Now that is one-to-one and onto.
 
pka said:
Why not do it directly and simply? \(\displaystyle f:\left[ {0,1} \right] \mapsto \left( {0,1} \right)\)
\(\displaystyle f(x) = \left\{ {\begin{array}{l,r} {\frac{1}{2},} & {x = 0} \\ {\frac{1}{3},} & {x = 1} \\ {\frac{1}{{n + 2}},} & {x = \frac{1}{n},\;n \geqslant 2} \\ {x,} & {else} \\ \end{array} } \right.\)

Now that is one-to-one and onto.
Click to expand...

[0,1) --->(0,1) is not correct or is it. Does that function include 0? How do I show it's 1-1 and onto?
 
kathrynag said:
Ok do I just not include the 1 part because 1 is not inluded?
Click to expand...
First, I turned the mapping around: \(\displaystyle \left[ {0,1} \right] \mapsto \left( {0,1} \right)\).
0 is mapped to one half, 1 is mapped to one third, each 1/n is mapped to 1/(n+2) and others are fixed.
 
pka said:
kathrynag said:
Ok do I just not include the 1 part because 1 is not inluded?
Click to expand...
First, I turned the mapping around: \(\displaystyle \left[ {0,1} \right] \mapsto \left( {0,1} \right)\).
0 is mapped to one half, 1 is mapped to one third, each 1/n is mapped to 1/(n+2) and others are fixed.
Click to expand...
Yeah, but right now I'm tryong to map [0,1) to (0,1) not [0,1] to (0,1).
 
It is a basic fact that a bijection from a set A to a set B has an inverse that is a bijection from B to A.
 
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