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 because onto means f(A)=B where the A=(0,1) and the B=(0,1]?