Advanced Calc. proof, about sets and intersection.

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Homework Statement


Prove that: A is a subset of B if and only if (A intersection B)=A


Homework Equations





The Attempt at a Solution

I tried proving the right side, that is

(A [tex]\cap[/tex] B)=A
For two sets to be equal then they have to be subsets of each other...so:

(A [tex]\cap[/tex] B) [tex]\subseteq[/tex] A and A [tex]\subseteq[/tex] (A [tex]\cap[/tex] B)
So if we assume an element x [tex]\in[/tex] (A[tex]\cap[/tex]B), then by definition, x[tex]\in[/tex]A and x [tex]\in[/tex]B. Thus we proved that (A[tex]\cap[/tex]B)[tex]\subseteq[/tex]A.

In not quite sure how to prove the opposite, because if x is an element of A, that doenst necessarily mean that x is an element of A[tex]\cap[/tex]B...so i need help with the rest of it..or if you got any other ideas on how to approach it.

Thank you,
Emira!
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Prove that: A is a subset of B if and only if (A intersection B)=A


Homework Equations





The Attempt at a Solution

I tried proving the right side, that is

(A [tex]\cap[/tex] B)=A
For two sets to be equal then they have to be subsets of each other...so:

(A [tex]\cap[/tex] B) [tex]\subseteq[/tex] A and A [tex]\subseteq[/tex] (A [tex]\cap[/tex] B)
So if we assume an element x [tex]\in[/tex] (A[tex]\cap[/tex]B), then by definition, x[tex]\in[/tex]A and x [tex]\in[/tex]B. Thus we proved that (A[tex]\cap[/tex]B)[tex]\subseteq[/tex]A.
Very good. That is exactly right!

In not quite sure how to prove the opposite, because if x is an element of A, that doenst necessarily mean that x is an element of A[tex]\cap[/tex]B...so i need help with the rest of it..or if you got any other ideas on how to approach it.

Thank you,
Emira![/QUOTE]
For the opposite, notice that you haven't used the hypothesis that A is a subset of B. If x is in A, then, because A is a subset of B it is also in B. Since it is in both A and B, it is in [itex]A\cap B[/itex]
Now you have to prove the implication the other way: If [itex]A\cap B\subseteq A[/itex] then [itex]A\subseteq B[/itex].
 

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