YAMT: Set Theory Proof

[DVK916]

If there is injection from A to B, and an injection from B to A then there must exist a bijection from A to B.

This isn't homework.

 

[DVK916]

Also I need the proof that does not use the Axiom Choice.

 

[DVK916]

Another way to look at this is.

Cardinality of A is less than or equal to the Cardinality B, and the Cardinality of B is less than or equal to the Cardinality of A. Then the Cardinality of A equals the Cardinality of B.

Prove this.

 

[yosuke188]

I can't really help you, but this sure sounds like aHW question to me

 

[BigJ]

Math kills people. If it wasn't for Math, people wouldn't have guns.

 

[DVK916]

Originally posted by: BigJ
Math kills people. If it wasn't for Math, people wouldn't have guns.

:laugh:

 

[DVK916]

Originally posted by: yosuke188
I can't really help you, but this sure sounds like aHW question to me

It isn't.

 

[Goosemaster]

onto is such a better word

 

[JujuFish]

Porkchop sandwiches.

 

[chambersc]

jesus christ man, let someone get a word in before you you hit the reply button. 4 replys in like 5minutes isn't necessary. there is an EDIT button for a reason.

 

[DVK916]

Where are all the great math people on ATOT.

 

[Jeff7181]

Originally posted by: DVK916
Another way to look at this is.

Cardinality of A is less than or equal to the Cardinality B, and the Cardinality of B is less than or equal to the Cardinality of A. Then the Cardinality of A equals the Cardinality of B.

Prove this.

Because I said so.

That was easy...

 

[stan394]

assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

 

[DVK916]

Originally posted by: stan394
assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

This isn't a valid proof. I don't remember why. But my professor explained this proof doesn't work.

 

[stan394]

Originally posted by: DVK916

Originally posted by: stan394
assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

This isn't a valid proof. I don't remember why. But my professor explained this proof doesn't work.

oh i suck

 

[Splork]

Originally posted by: DVK916

Originally posted by: stan394
assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

This isn't a valid proof. I don't remember why. But my professor explained this proof doesn't work.

WTF did you say it wasn't HW, when you know damn well it's HW?

-sp

 

[Goosemaster]

Originally posted by: Splork

Originally posted by: DVK916

Originally posted by: stan394
assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

This isn't a valid proof. I don't remember why. But my professor explained this proof doesn't work.

WTF did you say it wasn't HW, when you know damn well it's HW?

-sp

it might not be.

 

[yosuke188]

Originally posted by: DVK916

Originally posted by: yosuke188
I can't really help you, but this sure sounds like aHW question to me

It isn't.

Care to explain? You talk about a professor too... I'm just a little suspicious

 

[DVK916]

Originally posted by: Splork

Originally posted by: DVK916

Originally posted by: stan394
assume the contrary, that A != B. then there exists an element c which exists only in B but not A. since cardinality of A <= cardinality of B. with c, cardinality of A < cardinality of B, which violates the other problem statement.

This isn't a valid proof. I don't remember why. But my professor explained this proof doesn't work.

WTF did you say it wasn't HW, when you know damn well it's HW?

-sp

No he just asked us to look at this problem. Try to prove it on our own. But it isn't homework persay.

 

[Goosemaster]

<----knows what hte answer is but tryign to put it in the correct ofrm

 

[blustori]

Originally posted by: Goosemaster
<----knows what hte answer is but tryign to put it in the correct ofrm

which is the same thing as not knowing the answer.

 

[stan394]

i think i see what i am wrong. it's cardinality. not equivalent of the set.

 

[DVK916]

Originally posted by: stan394
i think i see what i am wrong. it's cardinality. not equivalent of the set.

Yea.

Also saying assume cardinality of A is greater than the cardinality of B, but that violates the assumption that the cardinality of B is greater than or equal to the cardinality of A. Next assume cardinality of A is lesser than the cardinality of B, but that violates the assumption that the cardinality of A is greater than or equal to the cardinality of B. Therefore the cardinality of A equals the cardinality of B. <---- Many would think this proof is valid, but it isn't.

 

[Goosemaster]

Originally posted by: blustori

Originally posted by: Goosemaster
<----knows what hte answer is but tryign to put it in the correct ofrm

which is the same thing as not knowing the answer.

Assuming that both A and B are sets for the function f


by definition if there exists an injection both ways, every Value in set A inputed into function F has a unique and corressponding value in Set B and every valin in set B has a unique and corressponding value in set A.

Therefore, there are no values that disprove an injection and by definition, that is the definition of a bijection


but I can't say that

 

[DVK916]

Originally posted by: Goosemaster

Originally posted by: blustori

Originally posted by: Goosemaster
<----knows what hte answer is but tryign to put it in the correct ofrm

which is the same thing as not knowing the answer.

Assuming that both A and B are sets for the function f


by definition if there exists an injection both ways, every Value in set A inputed into function F has a unique and corressponding value in Set B and every valin in set B has a unique and corressponding value in set A.

Therefore, there are no values that disprove an injection and by definition, that is the definition of a bijection


but I can't say that

You seem to be making a leap. I don't think you can go from that to that.