Prove (0,1)=(0,1] - Showing f is 1-1

[kathrynag]

I got everything except for proving that [0,1) is equivalent to (0,1], proving that (0,1) is equivalent to [0,1].
I'm a little confused on these 2 steps.

 

[kathrynag]

I got everything except for proving that [0,1) is equivalent to (0,1], proving that (0,1) is equivalent to [0,1].
I'm a little confused on these 2 steps.

[0,1) is equivalent to (0,1]
I still can't quite grasp this.
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.

 

[HallsofIvy]

"Equal" has a specific meaning: A= B if and only if the symbols "A" and "B" represent exactly the same thing. "Equivalent" varies: A is equivalent to B using some given equivalence relation. What equivalence relation are you using?

The obvious one would be "have the same cardinality" but you have assured us that is not the case!

Also, originally you were look at (0,1) and [0,1] and now you have changed to [0, 1) and (0, 1]. Is this a new problem?

If you are trying to prove they are the same cardinality, , looking at 1/n, 1/(n+1), 1/(n+2), etc. will do no good because the numbers you necessarily form a countable set and none of (0,1), [0,1], [0,1), nor (0,1] are countable.

Look at my original response, #8.

 

[kathrynag]

"Equal" has a specific meaning: A= B if and only if the symbols "A" and "B" represent exactly the same thing. "Equivalent" varies: A is equivalent to B using some given equivalence relation. What equivalence relation are you using?

The obvious one would be "have the same cardinality" but you have assured us that is not the case!

Also, originally you were look at (0,1) and [0,1] and now you have changed to [0, 1) and (0, 1]. Is this a new problem?

If you are trying to prove they are the same cardinality, , looking at 1/n, 1/(n+1), 1/(n+2), etc. will do no good because the numbers you necessarily form a countable set and none of (0,1), [0,1], [0,1), nor (0,1] are countable.

Look at my original response, #8.

No same problem. These are the last 2 steps. Proving [0,1) is equivalent to (0,1] and then proving (0,1) is equivalent to [0,1]

 

[HallsofIvy]

That's the hard way to do it! And, as I said before, what you are doing, looking at numbers of the form 1/n, is of no use because the integers are countable and none of these sets are countable.

As I said a long time ago, to define a one-to-one, onto function from [0,1] to (0,1)define f(x)= x for x any irrational number between 0 and 1, write all rational numbers in (0,1) in a list, r₁, r₂, etc. then define f(0)= r₁, f(1)= r₂, f(r_n)= r_n+2.

 

[kathrynag]

That's the hard way to do it! And, as I said before, what you are doing, looking at numbers of the form 1/n, is of no use because the integers are countable and none of these sets are countable.

As I said a long time ago, to define a one-to-one, onto function from [0,1] to (0,1)define f(x)= x for x any irrational number between 0 and 1, write all rational numbers in (0,1) in a list, r₁, r₂, etc. then define f(0)= r₁, f(1)= r₂, f(r_n)= r_n+2.

The hard way to do it is the way I was told to do it with proving [0,1) is equivalent to (0,1].

 

[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.

 

[kathrynag]

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]

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]

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]

Ok, I found something else. 2 sets A and B are equivalent iff there is a 1-1 function from A onto B.

 

[Vid]

After reading this topic, I am completely speechless.

 

[mutton]

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.

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]

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]

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 $$\mathbb{R}$$.

 

[kathrynag]

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?

 

[mutton]

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]

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]

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]

Ok I understand that.

 

[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

 

[kathrynag]

Ok, I'm going to 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 I am 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]

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]

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!

 

[kathrynag]

Also f(x)=1-x
and g:[0,1)-->[0,1]

 

[kathrynag]

Ok it's all figured out. They key was letting [0,1) was f(x)=x in (0,1) and f(0)=1. Then I just used transitivity.