Proving Set Theory: A⊆B Equivalence

[rbnphlp]

Hi,
Im only starting to learn about naive set theory from a book , so pardon me if my answer to the question is really obvious..

Prove that ..
A$$\subseteq$$B , if and only if A$$\cap$$B =A,if and only if A$$\cup$$B=B, if and only if A-B=empty set..

I was thinking of using venn diagrams to visualise it and it makes sense , however I do not know how to go on proving it..
thanks

 

[CompuChip]

Here are some things that are very very useful in general:

• To prove "X if and only if Y" first assume X and prove Y ("X => Y"), then assume Y and prove X ("Y => X").
• If A and B are two sets, to prove that A = B first show that $A \subseteq B$ and then that $B \subseteq A$.
• To prove that A is a subset of B, take any x in A and show that it is in B.

Such proofs with sets are usually very straightforward, they almost always proceed in more or less the same way. So the more you do, the easier it'll get.

Anyway, let's get to your question and start with the first one. Using my points above, I will first assume that $A \subseteq B$ and show that the intersection of A and B is equal to A. To show the latter, I only need to show that $A \subset A \cap B$, because the other inclusion is trivial.

So suppose that $A \subseteq B$. I want to show that $A \subset A \cap B$ so let x be an element in A. Because A is a subset of B, any element of A is also an element of B. So x is an element of B. From x in A and x in B, it is in the intersection of A and B. Since x was arbitrary, all elements of A lie in A intersect B, so A is a subset of A intersect B. By definition of intersection, A intersect B is a subset of A, so A intersect B = A.

Now try the converse implication and the other ones yourself, and try to remember the techniques from the proof above.

 

[rbnphlp]

Here are some things that are very very useful in general:

• To prove "X if and only if Y" first assume X and prove Y ("X => Y"), then assume Y and prove X ("Y => X").
• If A and B are two sets, to prove that A = B first show that $A \subseteq B$ and then that $B \subseteq A$.
• To prove that A is a subset of B, take any x in A and show that it is in B.

Such proofs with sets are usually very straightforward, they almost always proceed in more or less the same way. So the more you do, the easier it'll get.

Anyway, let's get to your question and start with the first one. Using my points above, I will first assume that $A \subseteq B$ and show that the intersection of A and B is equal to A. To show the latter, I only need to show that $A \subset A \cap B$, because the other inclusion is trivial.

So suppose that $A \subseteq B$. I want to show that $A \subset A \cap B$ so let x be an element in A. Because A is a subset of B, any element of A is also an element of B. So x is an element of B. .From x in A and x in B, it is in the intersection of A and B Since x was arbitrary, all elements of A lie in A intersect B, so A is a subset of A intersect B. By definition of intersection, A intersect B is a subset of A, so A intersect B = A.

Now try the converse implication and the other ones yourself, and try to remember the techniques from the proof above.

thank you makes perfect sense, except the underlined bit , can't x be the union of A and B , as x is in both A and B

 

[CompuChip]

Note, the definition of x being in the intersection is, that x is in both.
If x is in the union then it is in at least one of them (or both).

For example, if
A = {1, 2, 3}
B = {2, 3, 4}

Then 1 and 4 are in the union $A \cup B$ but not in the intersection $A \cap B$; 2 and 3 are both in the union and in the intersection

 

[rbnphlp]

Note, the definition of x being in the intersection is, that x is in both.
If x is in the union then it is in at least one of them (or both).

For example, if
A = {1, 2, 3}
B = {2, 3, 4}

Then 1 and 4 are in the union $A \cup B$ but not in the intersection $A \cap B$; 2 and 3 are both in the union and in the intersection

of course ... thanks again