prove (0,1)=[0,1]


by kathrynag
Tags: prove
mutton
mutton is offline
#37
Dec11-08, 09:33 AM
P: 179
kathrynag already stated a bijection between (0, 1) and (0, 1] in the very first post of this thread. (It doesn't take too much effort from that to conclude there is a bijection between (0, 1) and [0, 1].) Her problem is showing that it is one-to-one and onto.

I suggest practicing showing that simpler functions are bijections, such as f : R -> R defined by f(x) = 5 - x^3, so that you understand the form of the proof.
kathrynag
kathrynag is offline
#38
Dec11-08, 01:00 PM
P: 607
Quote Quote by mutton View Post
kathrynag already stated a bijection between (0, 1) and (0, 1] in the very first post of this thread. (It doesn't take too much effort from that to conclude there is a bijection between (0, 1) and [0, 1].) Her problem is showing that it is one-to-one and onto.

I suggest practicing showing that simpler functions are bijections, such as f : R -> R defined by f(x) = 5 - x^3, so that you understand the form of the proof.
f(a)=5-a^3
f(b)-5-b^3
5-x^3=5-b^3
a^3=b^3
a=b
Thus 1-1

y=5-x^3
-y+5=x^3
x=(-y+5)^1/3
x includes all real numbers. Thus onto.
Then how do i work around proving [0,1) is equivalent to (0,1]?
Like I thought about using f(x)=1-x because f(0)=1 and f(1)=0 but then that would be a mpping and that's not quite right.
kathrynag
kathrynag is offline
#39
Dec11-08, 11:06 PM
P: 607
Ok could I still use C=[0,1]
Then A=C-x element of C
B=C
but B=C-null set
null set is an element of C
so A=C- some elemnt of C=B?
kathrynag
kathrynag is offline
#40
Dec12-08, 10:19 PM
P: 607
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
kathrynag is offline
#41
Dec13-08, 12:25 PM
P: 607
Ok, I found something else. 2 sets A and B are equivalent iff there is a 1-1 function from A onto B.
Vid
Vid is offline
#42
Dec13-08, 12:34 PM
P: 420
After reading this topic, I am completely speechless.
mutton
mutton is offline
#43
Dec13-08, 12:49 PM
P: 179
Quote Quote by kathrynag View Post
Ok, I found something else. 2 sets A and B are equivalent iff there is a 1-1 function from A onto B.
This would make a lot more sense to us if equivalent is replaced with cardinally equivalent.

Quote Quote by kathrynag View Post
f(a)=5-a^3
f(b)-5-b^3
5-x^3=5-b^3
a^3=b^3
a=b
Thus 1-1

y=5-x^3
-y+5=x^3
x=(-y+5)^1/3
x includes all real numbers. Thus onto.
Then how do i work around proving [0,1) is equivalent to (0,1]?
Like I thought about using f(x)=1-x because f(0)=1 and f(1)=0 but then that would be a mpping and that's not quite right.
That is right, so use this method to show that the function you stated in your first post is a bijection between (0, 1) and (0, 1]. The proof will be longer because that function is case-defined.

You already have f(x) = 1 - x, a bijection between [0, 1) and (0, 1], which is easier to show.

If you also have a bijection between [0, 1) and [0, 1], then that implies (0, 1) and [0, 1] are cardinally equivalent. If you want a single bijection between (0, 1) and [0, 1], think about how you can "combine" the 3 functions.
kathrynag
kathrynag is offline
#44
Dec13-08, 01:41 PM
P: 607
Quote Quote by mutton View Post
This would make a lot more sense to us if equivalent is replaced with cardinally equivalent.



That is right, so use this method to show that the function you stated in your first post is a bijection between (0, 1) and (0, 1]. The proof will be longer because that function is case-defined.

You already have f(x) = 1 - x, a bijection between [0, 1) and (0, 1], which is easier to show.

If you also have a bijection between [0, 1) and [0, 1], then that implies (0, 1) and [0, 1] are cardinally equivalent. If you want a single bijection between (0, 1) and [0, 1], think about how you can "combine" the 3 functions.
Well, the book says nothing about cardinally equivalence, so i would assume my professor would not expect me to use that term.
Ok, so (0,1) and (0,1]
For f(x)=x, we get 0 and 1
For f(1/n)=1/(n-1), we get {1/1/2,1/1,1/4....}
Thus, (0,1]


[0, 1) and [0, 1]
For this I use {0}U(0,1) because I already know (0,1) is a bijection to (0,1]. Thus we have a bijection to [0,1].

Ok, for (0,1) equivalent to [0,1].
Let f(x)=x for all elemnt not of the form 1/n
Then we get the points (0,1)
Thne for all other points let f(x)={0}U 1/(n-1)
mutton
mutton is offline
#45
Dec13-08, 01:59 PM
P: 179
Quote Quote by kathrynag View Post
Well, the book says nothing about cardinally equivalence, so i would assume my professor would not expect me to use that term.
Ok, so (0,1) and (0,1]
For f(x)=x, we get 0 and 1
For f(1/n)=1/(n-1), we get {1/1/2,1/1,1/4....}
Thus, (0,1]
You don't get 1 from f(x) = x.

Also, what about all other points not in {0, 1, 1/2, 1/3, 1/4, ...}?

[0, 1) and [0, 1]
For this I use {0}U(0,1) because I already know (0,1) is a bijection to (0,1]. Thus we have a bijection to [0,1].
Right idea, but you need to say it precisely because [tex]\{0\} \cup (0, 1) \ne (0, 1][/tex]. Do you mean given a bijection [tex]f : (0, 1) \rightarrow (0, 1][/tex], define [tex]g : [0, 1) \rightarrow [0, 1][/tex] by g(0) = 0, and g(x) = f(x) for [tex]x \in (0, 1)[/tex]?

Thne for all other points let f(x)={0}U 1/(n-1)
This is not an element of [tex]\mathbb{R}[/tex].
kathrynag
kathrynag is offline
#46
Dec13-08, 02:34 PM
P: 607
Quote Quote by mutton View Post
You don't get 1 from f(x) = x.

Also, what about all other points not in {0, 1, 1/2, 1/3, 1/4, ...}?



Right idea, but you need to say it precisely because [tex]\{0\} \cup (0, 1) \ne (0, 1][/tex]. Do you mean given a bijection [tex]f : (0, 1) \rightarrow (0, 1][/tex], define [tex]g : [0, 1) \rightarrow [0, 1][/tex] by g(0) = 0, and g(x) = f(x) for [tex]x \in (0, 1)[/tex]?



This is not an element of [tex]\mathbb{R}[/tex].
Yeah, but from f(x)=x I get all points not of the form 1/n.
All other points not in {0,1,1/2,1/3,1/4....} are in f(x)=x


Yeah, i guess setting g(0)=0 and g(x)=f(x) would work.

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) and let g(1)=0?
mutton
mutton is offline
#47
Dec13-08, 03:01 PM
P: 179
Quote Quote by kathrynag View Post
Yeah, but from f(x)=x I get all points not of the form 1/n.
All other points not in {0,1,1/2,1/3,1/4....} are in f(x)=x
Close. 0 is not in the range of f. I mentioned 0 only because you mentioned 0.

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) and let g(1)=0?
g(1) cannot be defined because the domain of g is (0, 1).
kathrynag
kathrynag is offline
#48
Dec13-08, 03:18 PM
P: 607
Quote Quote by mutton View Post
Close. 0 is not in the range of f. I mentioned 0 only because you mentioned 0.



g(1) cannot be defined because the domain of g is (0, 1).
Ok, 0 is not even in f.

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)=?
HallsofIvy
HallsofIvy is offline
#49
Dec13-08, 03:39 PM
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If If f:(0,1)-->(0,1] then there is NO "x" left over to give f(x)= 0. That's the whole point.
kathrynag
kathrynag is offline
#50
Dec13-08, 03:44 PM
P: 607
Ok I understand that.
kathrynag
kathrynag is offline
#51
Dec13-08, 06:12 PM
P: 607
Quote Quote by kathrynag View Post
Ok, 0 is not even in f.

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)=?
Is that the right idea or not?
I just am unsure of what x value to plug into g(x)=0
kathrynag
kathrynag is offline
#52
Dec13-08, 07:12 PM
P: 607
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)=?
kathrynag
kathrynag is offline
#53
Dec14-08, 10:40 AM
P: 607
Ok, do I use a composition?
Let f:(0,1)-->(0,1]
Let g:=[0,1)-->[0,1]
Let g(x)=1-x
Then h(x)=g(f(x))
kathrynag
kathrynag is offline
#54
Dec14-08, 11:52 AM
P: 607
Ok then I would also need to define g(0) as 0 right because we get:
h(x)=g([0,1]).
No never mind because I should be doing
h(x)=f(g(x)
=f([0,1]
f(0)=1 and f(1)=0
Please let me know if this is right!


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