• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

YAMT: Set Theory Proof

D

DVK916

Banned
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.
 
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.
 
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.
 
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.
Click to expand...

Because I said so.

That was easy...
 
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.
 
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.
Click to expand...

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

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

oh i suck 😛
 
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.
Click to expand...

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

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

-sp

 
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.
Click to expand...

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

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

-sp
Click to expand...

it might not be.
 
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.
Click to expand...

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

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

-sp
Click to expand...


No he just asked us to look at this problem. Try to prove it on our own. But it isn't homework persay.
 
Originally posted by: stan394
i think i see what i am wrong. it's cardinality. not equivalent of the set.
Click to expand...


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

which is the same thing as not knowing the answer. 🙂
Click to expand...


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
 
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
Click to expand...

which is the same thing as not knowing the answer. 🙂
Click to expand...


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
Click to expand...

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

 
You must log in or register to reply here.

TRENDING THREADS

Back
Top