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

1. Jun 25, 2012

### 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.

2. Jun 25, 2012

### micromass

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

3. Jun 25, 2012

### jmjlt88

You're right!! :)

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

4. Jun 25, 2012

### micromass

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

5. Jun 25, 2012

### 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 = bk for some integer k. Hence, <a>=<bk>, 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

6. Jun 25, 2012

### micromass

Staff Emeritus
Yeah, that seems like a perfectly valid proof!!