# Prove (0,1)=[0,1]

by kathrynag
Tags: prove
 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.
P: 607
 Quote by mutton 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.
 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?
 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]
 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.
 P: 420 After reading this topic, I am completely speechless.
P: 179
 Quote by kathrynag 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 by kathrynag 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.
P: 607
 Quote by mutton 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)
P: 179
 Quote by kathrynag 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 $$\{0\} \cup (0, 1) \ne (0, 1]$$. Do you mean given a bijection $$f : (0, 1) \rightarrow (0, 1]$$, define $$g : [0, 1) \rightarrow [0, 1]$$ by g(0) = 0, and g(x) = f(x) for $$x \in (0, 1)$$?

 Thne for all other points let f(x)={0}U 1/(n-1)
This is not an element of $$\mathbb{R}$$.
P: 607
 Quote by mutton 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 $$\{0\} \cup (0, 1) \ne (0, 1]$$. Do you mean given a bijection $$f : (0, 1) \rightarrow (0, 1]$$, define $$g : [0, 1) \rightarrow [0, 1]$$ by g(0) = 0, and g(x) = f(x) for $$x \in (0, 1)$$? This is not an element of $$\mathbb{R}$$.
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?
P: 179
 Quote by kathrynag 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).
P: 607
 Quote by mutton 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)=?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 If If f:(0,1)-->(0,1] then there is NO "x" left over to give f(x)= 0. That's the whole point.
 P: 607 Ok I understand that.
P: 607
 Quote by kathrynag 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
 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)=?
 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))
 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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