# Math Help - Proof: Section- Finite and Infinite Sets

1. ## Proof: Section- Finite and Infinite Sets

Hello, this is for my Advanced Calculus Class.

If A, B, and C are sets, then prove...

a) A ~ A
b) A ~ B if and only if B ~ A
c) A ~ B and B ~ C implies A ~ C

I need to prove each one, I tried working on each but I just end up getting lost or messing it up and it doesn't make sense. Please prove them.

2. Originally Posted by Mush89
If A, B, and C are sets, then prove...
a) A ~ A
b) A ~ B if and only if B ~ A
c) A ~ B and B ~ C implies A ~ C
This is a perfect example of your needing to include more detail when posting. It may surprise you to know that $A\sim A$ may mean anything. I taught advanced for thirty years. I have never seen that notation.
So tell us the meaning and definition of all relevant terms.

3. "~" means equivalent, or equipotent. It's used all over the Math books I've seen.

4. Originally Posted by Mush89
"~" means equivalent, or equipotent. It's used all over the Math books I've seen.
How does your textbook define equipotent?
Again there are several different approaches to this topic.
If we have to guess, it may just confuse you more.
Please be detailed when posting questions.

5. Originally Posted by Mush89
Hello, this is for my Advanced Calculus Class.

If A, B, and C are sets, then prove...

a) A ~ A (symmetry)
b) A ~ B if and only if B ~ A (reflexivity)
c) A ~ B and B ~ C implies A ~ C (transitivity)

I need to prove each one, I tried working on each but I just end up getting lost or messing it up and it doesn't make sense. Please prove them.
I added items in parentheses to above quote.

Shilov uses $\sim$ to denote equivalence, and defines equivalence to be a one-to-one correspondence between sets. In that case, symmetry and reflexivity are obvious, and for transitivity:
Corresponding to a in A there is a unique b in B and corresponding to b in B there is a unique c in C so corresponding to a in A there is a unique c in C. And then the other way around.

6. Originally Posted by Hartlw
I added items in parentheses to above quote.
a) A ~ A (symmetry)
b) A ~ B if and only if B ~ A (reflexivity)
c) A ~ B and B ~ C implies A ~ C (transitivity)
Shilov uses $\sim$ to denote equivalence, and defines equivalence to be a one-to-one correspondence between sets.
Hartlw, you have once again jumped in without getting it right.
You have the two terms in red above backwards.
Moreover, you do not know that the OP have with Shilov's way of defining this relation.

7. This proof is for my Advanced Calculus class. It's in a section called Finite and Infinite Sets. I need to prove a, b, c if A, B, and C are sets.

8. Originally Posted by Mush89
This proof is for my Advanced Calculus class. It's in a section called Finite and Infinite Sets. I need to prove a, b, c if A, B, and C are sets.
OK. You wrote that $A\sim B$ in your text material means that $A\text{ is equipotent to }B$.
Now tell us how the textbook defines equipotent.
At this point we are still just guessing.

9. It says, "Two sets, A and B, are equivalent, or equipotent, denoted by A ~ B, if and only if a bijection exists from A onto B. And equipotent meaning they have the same cardinality.

10. Originally Posted by Mush89
It says, "Two sets, A and B, are equivalent, or equipotent, denoted by A ~ B, if and only if a bijection exists from A onto B. And equipotent meaning they have the same cardinality.
Thank you. Had that been included to begin with, you would have gotten real help sooner;.
$f:x\mapsto x$ the identity mapping is a bijection of any set to itself. Thus, $A\sim A.$

If $f:A\to B$ is a bijection then $f^{-1}:B\to A$ is also a bijection. Thus we have symmetry.

If $f:A\to B~\&~g:B\to C$ are both bijections then $g\circ f:A\to C$ is also a bijection.

11. Originally Posted by Plato
Thank you. Had that been included to begin with, you would have gotten real help sooner;.
$f:x\mapsto x$ the identity mapping is a bijection of any set to itself. Thus, $A\sim A.$

If $f:A\to B$ is a bijection then $f^{-1}:B\to A$ is also a bijection. Thus we have symmetry.

If $f:A\to B~\&~g:B\to C$ are both bijections then $g\circ f:A\to C$ is also a bijection.
This was already done in above post by Hartlw. From Wofram Math World:

"Two sets A and B are said to be equipollent iff there is a one-to-one correspondence (i.e., a bijection) from A onto B (Moore 1982, p. 10; Rubin 1967, p. 67; Suppes 1972, p. 91)."

12. Originally Posted by Hartlw
This was already done in above post by Hartlw. From Wofram Math World:

"Two sets A and B are said to be equipollent iff there is a one-to-one correspondence (i.e., a bijection) from A onto B (Moore 1982, p. 10; Rubin 1967, p. 67; Suppes 1972, p. 91)."
Plato's point was that there are several different (equivalent) ways to define equipotency, and without knowing which way the author is referring to, we can not really help him.

13. Originally Posted by Defunkt
Plato's point was that there are several different (equivalent) ways to define equipotency, and without knowing which way the author is referring to, we can not really help him.
The only one I could find wrt two sets A and B is one-to-one onto. Do you know of any others?

14. Yes - an equivalent definition is that there is an injection $f: A \to B$ and an injection $g: B \to A$.

15. Originally Posted by Defunkt
Yes - an equivalent definition is that there is an injection $f: A \to B$ and an injection $g: B \to A$.
From Wolfram Math World (google "injection"):

"In other words, is an injection if it maps distinct objects to distinct objects. An injection is sometimes also called one-to-one."

However, the conditions a), b), and c) of OP can only be satisfied if it is one-to-one-onto, which I used in my proof, ie, we are still talking about the same thing, one-to-one onto.

Speaking of my original proof, I identified a) and b) of OP as symmetry and reflexivity, when they are conventionally identified the other way around, except by Shilov (pg 30), which I happened to have at hand and because I can't remember which is which and $A \sim A$ looks symmetric to me- I guess it makes more sense that way to us Russians. Technically, and from a purist point of view, having defined my terms my proof was correct within the context of the definition, and, btw, also for the reverse definition, and it really didn't depend on how I happened to identify conditions a) and b) anyhow. Its a triviality which produced a very harsh response from Plato, and a warning message from the powers that be.

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