# Homework Help: Prove (0,1)=[0,1]

1. Dec 2, 2008

### kathrynag

1. The problem statement, all variables and given/known data
prove (0,1)=[0,1]
Define f;(0,1)-->R as follows.
For n$$\in$$natural numbers, n$$\geq$$2, f(1/n)=1/(n-1) and for all other x$$\in$$(0,1), f(x)=x.

2. Relevant equations

3. 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.

2. Dec 2, 2008

### 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.

3. Dec 2, 2008

### kathrynag

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

4. Dec 2, 2008

### mutton

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

5. Dec 2, 2008

### 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)

6. Dec 2, 2008

### kidmode01

Ah that makes more sense mutton.

7. Dec 2, 2008

### kathrynag

It's a real analysis proof

8. Dec 3, 2008

### 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".

9. Dec 3, 2008

### 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)

10. Dec 3, 2008

### mutton

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

11. Dec 3, 2008

### kathrynag

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

12. Dec 3, 2008

### rochfor1

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

13. Dec 3, 2008

### 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.

14. Dec 3, 2008

### 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.

15. Dec 3, 2008

### kidmode01

Haha k I guess it wont 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.

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

16. Dec 3, 2008

### 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.)

17. Dec 3, 2008

### kidmode01

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

18. Dec 7, 2008

### 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.

So, we use im 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.

19. Dec 7, 2008

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

$$[0,1] = (0,1) \cup \{0,1\}$$

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?

20. Dec 7, 2008

### kathrynag

We have done nothing with cardinality.

21. Dec 7, 2008

### 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.

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.

22. Dec 8, 2008

### kathrynag

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

23. Dec 8, 2008

### mutton

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

24. Dec 8, 2008

### kathrynag

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

25. Dec 8, 2008

### mutton

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