prove (0,1)=[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

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

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

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.

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]


[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

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].

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


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

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).

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)=?

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

Is that the right idea or not?
I just am unsure of what x value to plug into g(x)=0