Set Theory

1. No it wouldn't. First of all, you haven't said what [itex]\overline{A}[/itex] means (is it complement?). Second of all, I assume that's a set, so [itex]\overline{A} \not\in U[/itex] doesn't make sense (did you mean [itex]\not\subset[/itex]?). Actually the question is quite easy, try to make a picture. You can construct a counterexample by for example looking at subsets of the natural numbers (e.g. [tex]A = \{0, 1, 2, 3\}, B = \{3, 5\} \cup \{ \text{ even numbers } \}, C = \mathbb{N}[/tex]).

For 2, it might help to draw a diagram again.

For a statement like 3, generally one takes an element in the left hand side and shows that it is also in the set on the right hand side and conversely to show that the LHS is contained in the RHS and vice versa.

 

Actually, that makes no sense at all. You are given that U is the universal set so both A and [itex]\bar{A}[/itex] are subsets of U. That has nothing to do with [itex]\bar{A}[/itex] being a member of U. In any case, you are only asked to give a counterexample. Suppose B= {a,b,c,d,e,f}, C= {d,e,f,g, i, j}. Can you find A that is a both a subset of B and a subset of C?

Well, I'll give you a start. (2) says, "[tex](A\cap B) \cup C = (A \cap (B \cup C))[/tex] if and only if [tex]C \subseteq A[/tex]
That is an "if and only if" statement so you need to prove 2 things:
a) If [tex](A\cap B) \cup C = (A \cap (B \cup C))[/tex] then [tex]C \subseteq A[/tex]

b) If [tex]C \subseteq A[/tex] then (A\cap B) \cup C = (A \cap (B \cup C))[/tex]
and the standard way to prove "[itex]A= B[/itex]" is to say "If [itex]a\in A[/itex]" and show that [itex]a\in B[/itex]. That is, assuming that a satisfies whatever conditions define A, show that it must satisfy whatever conditions satisfy B.

To show [tex]C \subseteq A[/tex], start by saying "if [itex]x \in C[/itex] and use the fact that [tex]](A\cap B) \cup C = (A \cap (B \cup C))[/tex] to show [itex]x \in A[/itex].

To show [tex]A \subseteq C[/tex], start by saying "if [itex]x \in A[/itex] and use the fact that [tex]](A\cap B) \cup C = (A \cap (B \cup C))[/tex] to show [itex]x \in C[/itex].

Then do it the other way around.

Same thing. Don't write set operations like that. Start by saying "if [tex]x \in (A- B)- C[/tex] and then show that [tex]x \in (A- C)- (B- C)[/itex]. Using the definitions of those set operations, of course.

 

Those seem like very complicated examples for 1).

Let's "just do it". A is a subset of B. OK, so let A={a}, and let B=A. Does that work? No.

We're going to have to make B bigger. How about B={a,b}. Now I need to find a C which contains A, and doesn't contain B. Well, that's easy: C=A.

I also think the advice for 2) is overly long. We are surely allowed to use the identity

(A n B) u C = (A u C) n (B u C)

which makes one direction trivial.

Now, what does it mean for C not to be a subset of A? It means that there is a c in C that is not in A. Which proves the other direction.

 

If A and B were necessarily equal then you would write A= B! If A and B were necessarily not equal, then you would write [tex]A\subset B[/tex]. [tex]A\subseteq B[/tex] means that A may be equal to B or may just be a subset.

 

Sometimes the notation [itex]A \subset B[/itex] is also used for "A is either a proper subset of B, or A and B are equal" and "A is a proper subset of B, they are not equal" is explicitly written as [itex]A \subsetneq B[/itex]. It depends on the author (and most authors will -- and all of them should -- explain their conventions somewhere at the start).

 

I don't quite see what you proved. You have to show that:
If [itex]C \subseteq A[/itex] then [itex](A \cap B) \cup C = A \cap (B \cup C)[/itex] --- so if [itex]C \subseteq A[/itex] then [itex]x \in (A \cap B) \cup C \implies x \in A \cap (B \cup C)[/itex] and [itex]x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup C [/itex]

And you have to show that if [itex](A \cap B) \cup C = A \cap (B \cup C)[/itex], then [itex]x \in C[/itex] implies [itex]x \in A[/itex] (the other way around is not needed, and might not even be true).

 

If x is contained in B or C it does not follow that x is contained in C!

 

No, no one said anything like that. If x is an element of [itex](A\cap B)\cup C[/itex] then x is an element of [itex]A\cap B[/itex] or x is a member of C. For example, suppose A= {a, b}, B= {b, c} and C= {p, q}. Then [tex]A\cap B[/itex]= {b} and so [tex](A\cap B)\cup C}[/tex]= {b, p, q}. If x= p or x= q, then it is true that x is an element of [itex](A\cap B)\ cup C[/itex] but x is not an element of [itex] A \cap (B \cup C) [/itex] .

I have know idea what you mean by "by Union of [itex](A\cap B)\cup C[/itex]". Are you clear on the definitions of "union" and "intersection"? If x is a member of C, then, yes, it is a member of [itex](A\cap B)\cup C[/itex] but that tells you nothing at all about whether x is or is not a member of [itex]A\cap B[/itex].
I think you need to go back and review the definitions.

 

You mean by definition of ...

Unless you have additional information you don't- it's not necessarily true. What information do you have about A, B, and C?

You are getting yourself confused. In order to prove [tex]C \subseteq A[/tex], yes you should start with "if x is a member of C". Now use the fact that [tex](A\cap B) \cup C = (A \cap (B \cup C))[/tex]. If x is in C, then it is in [tex](A\cap B)\cup C[/tex] which mean it is in [tex](A\cap (B\cup C))[/tex] (because they are equal) which, in turn, means x is in ?

 

Look, this really is trivial, and shouldn't have taken this long.

We know that [itex](A\cap B)\cup C)=(A\cup C)\cap (B\cup C)[/itex], so that is one direction proved. Do you see that? Conversely, if A is not a subset of C....