- #26

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

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- #26

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

- #27

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y=[tex]\frac{1}{n-1}[/tex]

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

- #28

HallsofIvy

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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

- #29

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

- #30

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- #31

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I'm a little confused on these 2 steps.

- #32

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[0,1) is equivalent to (0,1]

I'm a little confused on these 2 steps.

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.

- #33

HallsofIvy

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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

Look at my original response, #8.

- #34

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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]

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 youaretrying 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.

- #35

HallsofIvy

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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, r

- #36

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The hard way to do it is the way I was told to do it with proving [0,1) is equivalent to (0,1].

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, r_{1}, r_{2}, etc. then define f(0)= r_{1}, f(1)= r_{2}, f(r_{n})= r_{n+2}.

- #37

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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.

- #38

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f(a)=5-a^3

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(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.

- #39

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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?

- #40

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Now I'm having trouble finding a similar function for (0,1) equivalent to [0,1]

- #41

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Ok, I found something else. 2 sets A and B are equivalent iff there is a 1-1 function from A onto B.

- #42

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After reading this topic, I am completely speechless.

- #43

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This would make a lot more sense to us ifOk, I found something else. 2 sets A and B are equivalent iff there is a 1-1 function from A onto B.

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.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.

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.

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- #44

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Well, the book says nothing about cardinally equivalence, so i would assume my professor would not expect me to use that term.This would make a lot more sense to us ifequivalentis replaced withcardinally 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.

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)

- #45

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You don't get 1 from f(x) = x.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]

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]?[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].

This is not an element of [tex]\mathbb{R}[/tex].Thne for all other points let f(x)={0}U 1/(n-1)

- #46

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Yeah, but from f(x)=x I get all points not of the form 1/n.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].

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?

- #47

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Close. 0 is not in the range of f. I mentioned 0 only because you mentioned 0.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

g(1) cannot be defined because the domain of g is (0, 1).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?

- #48

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Ok, 0 is not even in f.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).

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)=?

- #49

HallsofIvy

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If If f:(0,1)-->(0,1] then there is NO "x" left over to give f(x)= 0. That's the whole point.

- #50

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Ok I understand that.

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