# Prove (0,1)=[0,1]

## Homework Statement

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.

## 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.

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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.

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

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

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)

Ah that makes more sense mutton.

It's a real analysis proof

HallsofIvy
Homework Helper
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".

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)

Oops, i guess it is (0,1)=[0,1].
No, it's not. Get your notation right.

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].

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

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.

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.

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.

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.)

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

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.

Homework Helper
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?

We have done nothing with cardinality.

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.

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

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

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

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