For sets A,B, if I can that show |A|=|B| and A is a subset of B, then A=B?

[jmjlt88]

[Note: If this is posted in the wrong forum, I'm very sorry. It is directly related to a textbook question.]

This may be a silly question.

I know that I can prove two sets to be equal by showing that they are subsets of each other. But, what if I have that two sets have the same cardinality and that one set is a subset of the other. Is this enough to show that the sets are indeed equal?

Thanks.

 

[micromass]

Think about $A=\mathbb{N}$ and $B=\mathbb{Z}$.

 

[jmjlt88]

You're right! :)

But, I forgot to mention that they are finite. :(

 

[micromass]

OK, then it is indeed true. Do you need to prove this?? It's pretty obvious, but the proof might be quite complicated.

 

[jmjlt88]

Suppose a ε <b>. Then <a> = <b> iff a and b have the same order.

Proof [just the converse] :
Let G be a group. Let a,b be elements of G. Suppose a ε <b> and suppose a and b have the same order. Let n=ord(a)=ord(b). Then, the cyclic groups generated by a and b have the same order. That is, n=|<a>|=|<b>|. Now, from our assumption, a ε <b> and, then, since <b> consists of all powers of b, a = b^k for some integer k. Hence, <a>=<b^k>, which is a subset of <b>. Then since <a> is a subset of <b> and they have the same cardinality, they must be equal. Hence, <a>=<b>, our desired result. QED

 

[micromass]

Suppose a ε <b>. Then <a> = <b> iff a and b have the same order.

Proof [just the converse] :
Let G be a group. Let a,b be elements of G. Suppose a ε <b> and suppose a and b have the same order. Let n=ord(a)=ord(b). Then, the cyclic groups generated by a and b have the same order. That is, n=|<a>|=|<b>|. Now, from our assumption, a ε <b> and, then, since <b> consists of all powers of b, a = b^k for some integer k. Hence, <a>=<b^k>, which is a subset of <b>. Then since <a> is a subset of <b> and they have the same cardinality, they must be equal. Hence, <a>=<b>, our desired result. QED

Yeah, that seems like a perfectly valid proof!