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

[kathrynag]

Homework Statement

prove (0,1)=[0,1]
Define f;(0,1)-->R as follows.
For n\innatural numbers, n\geq2, f(1/n)=1/(n-1) and for all other x\in(0,1), f(x)=x.

Homework Equations

The Attempt at a Solution

I know I need to start proving f is a 1-1 function from (0,1) into (0,1]. The problem si getting started.
I know for 1-1 f(y)=f(z) implies y=z.

 

[kidmode01]

Prove (0,1) = [0,1] ?
Or perhaps disprove? Remember, intervals are just sets. To prove 2 sets are equal, you must prove each set is a subset of the other.

(0,1) is a subset of [0,1], but [0,1] is not a subsets of (0,1) since 0 and 1 are not in (0,1). So It is not true that (0,1)=[0,1]

As for your second question. You want to prove F is 1-1? A function F:A->B is 1-1 if for all a and a' in A, f(a) = f(a') implies a = a'

So suppose f(a) = f(a') ... now this is all you have to work with, evaluate what f(a) and f(a') is.

 

[kathrynag]

Well sorry it's to prove [0,1]=(0,1), but it is possible.

 

[mutton]

It seems what you really want to prove is |(0,1)| = |[0,1]|, where | | represents cardinality.

 

[kidmode01]

The intervals we're talking about are sets of real numbers and it doesn't matter what order you put the equality. Either way, [0,1] is not a subset of (0,1). Why?

If [0,1] were equal to (0,1) then [0,1] would be a subset of (0,1). That means that every number in [0,1] would be inside (0,1) but that is clearly not the case since the boundary points of [0,1] are not included in (0,1).

Please show me proof that [0,1] = (0,1)

 

[kidmode01]

Ah that makes more sense mutton.

 

[kathrynag]

It's a real analysis proof

 

[HallsofIvy]

The first thing you need to do is read the problem carefully!

You have twice now told us that the problem is to prove that [0,1]= (0,1).

You can't- it's not true.

You also said "I know I need to start proving f is a 1-1 function from (0,1) into (0,1]", which would NOT prove that [0,1]= (0,1). It would prove that the cardinality of (0, 1) is the same as the cardinality of [0,1]. That is completely different from saying "(0,1)= [0,1]".

Try mapping the irrational numbers to themselves, then look at the rationals only. Since they are countable, they can be put into a sequence- and then "shifted".

 

[kathrynag]

Oops, i guess it is (0,1)=[0,1].
Ok, then evaluate for a and a'.
f(1/a)=1/(a-1))
f(1/a')=1/(a'-1)
1/(a-1)=1/(a'-1)

 

[mutton]

Oops, i guess it is (0,1)=[0,1].

No, it's not. Get your notation right.

 

[kathrynag]

No, it's not. Get your notation right.

yes it is. I just checked my book and my book says to prove (0,1)=[0,1].

 

[rochfor1]

Definitely not true. (0,1) doesn't include 0. [0,1] does.

 

[kidmode01]

This is frustrating haha so this will be my last post on this thread :) Using standard interval notation (0,1) is not equal to [0,1]. I think you should do a little more research into set theory and what an interval actually is (It's a set) and then you might understand what you are trying to tell us. Either you are not including enough information, or there is a typo in your book. There is no convincing us otherwise.

 

[kathrynag]

Ok, sorry it's frustrating me too because I've always been told that (0,1) doesn't include 0 and 1, but [0,1] does, but now I have to prove these 2 are equal. It's hard to grasp.

 

[kidmode01]

Haha k I guess it won't be my last post :) You've been told correctly. (0,1) does not include 0 and 1 but [0,1] does. Both of those intervals are sets. Sets are just collections of numbers. Are the sets {1,2} and {1,2,3} equal? What about the sets {1} and {2}? Of course not! By the same reasoning [0,1] is not equal to (0,1) because (0,1) doesn't have 0 and 1.

[Edit]

Well more generally, a set is a collection objects but in our case we'll be using real numbers.

 

[mutton]

You are not proving that they are equal. You are proving that they have the same number of elements.

Just check that your function is a bijection between the two sets. (Read HallsofIvy's last paragraph.)

 

[kidmode01]

Oh geez, I should have been a little more careful with my last post. But yes, listen to mutton and HallsofIvy.

 

[kathrynag]

I did a little more...


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 elemts of(0,1), f(x)=x.

I have some steps to follow, proving that f is a 1-1 function from (0,1) into (0,1] and proving that f is a function from (0,1) onto (0,1] are the first 2 steps.
I know how to prove 1-1.
Let a=1/n and b=1/m
f(a)=1/n-1
f(b)=1/m-1
1/n-1=1/m-1
m-1/n-1=1
m-1=n-1
m=n and thus 1-1.
I see that but I'm unsure how to prove that it goes to (0,1] because of the 1. I know 1/2-->1, but I assume this doesn't really prove it.

I'm unsure about proving onto.
So, we use I am f.
y=1/n-1
y(n-1)=1
n-1=1/y
n=1/y +1
Let a= x and b=z
n=1/x+1 and n=1/z+1
1/x+1=1/z+1
1/x=1/z
1=x/z
z=x, thus onto.
My problem is proving for the specific intervals.

 

[statdad]

It has to be to show the sets have the same cardinality. Notice that

<br /> [0,1] = (0,1) \cup \{0,1\}<br />

and that (0,1) and \{0,1\} are disjoint. You should know the cardinality of both [0,1] and (0,1), so what can you conclude?

 

[kathrynag]

We have done nothing with cardinality.

 

[mutton]

For 1-1, you want to show that if f(a) = f(b), then a = b. You have done this only when a and b are of the form 1/n. What about all the other possible values of a and b?

For onto, you want to show that if y is in (0, 1], then there is an x in (0, 1) such that f(x) = y. Again, there should be more than 1 case for y because of the way f is defined.

We have done nothing with cardinality.

But it is exactly what you are doing, even if you don't recognize the name. When you show that there is a bijection between 2 sets, you show that they have the same cardinality.

 

[kathrynag]

Could someone at leats let me know if I did the right thing for proving onto?

 

[mutton]

No, your notation is flawed so it's not even clear what you are doing.

 

[kathrynag]

No, your notation is flawed so it's not even clear what you are doing.

f(1/n)=\frac{1}{n-1}
y=\frac{1}{n-1}
y(n-1)=1
n-1=1/y
n=\frac{1}{y}+1

n=\frac{1}{a}+1
n=\frac{1}{b}+1
\frac{1}{a}+1=\frac{1}{b}+1
\frac{1}{a}=\frac{1}{b}
1=\frac{a}{b}
b=a
Thus onto

 

[mutton]

How does that show f is onto? Did you read what I wrote above?

 

[kathrynag]

Yeah I read what you wrote above, but I'm unsure how to fix it.

 

[kathrynag]

f(1/n)=\frac{1}{n-1}
y=\frac{1}{n-1}
y(n-1)=1
yn-y=1
yn=1+y
y=1/n+y/n
y-y/n=1/n
y(1-1/n)=1/n

 

[HallsofIvy]

I have no idea what you are trying to do. You have repeatedly said that you are trying to prove "[0, 1]= (0, 1)" which can't be proved- it isn't true.

You have repeatedly said this problem has nothing to do with "cardinality" but are trying to find a "one-to-one, onto" function between the two sets which, if there is one, would prove cardinality not equality.

 

[descendency]

I think the problem is lazy notation here.

The OP doesn't mean [0,1] = (0,1) or else the simple counter-example 1 \in [0,1] but 1 \notin (0,1) proves it to be false; they likely mean [0,1] \approx (0,1). (A is equivalent to B)

The proof for this is to prove their cardinality is the same. The proof of equivalence simply requires a one-to-one correspondence between (0,1) and [0,1]. (as has been said before, this is the proof for cardinality).

 

[kathrynag]

Ok, so does for the function f(1/n)={1.1/2,1/3,1/4,1/5...} imply onto becuase onto means f(A)=B where the A=(0,1) and the B=(0,1]?

 

[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.