There Exist Sets A And B Such That Int A

Show that if f: A->B and E,F are subsets of A, then f(E U F) = f(E) U f(F).?

Let's break it up into two parts and put it all together.

First some preliminaries.

I'll take it for granted that f(E) and f(F) exist, since you've already shown they do (yes, that is enough). Naturally f(E U F) must also exist since everything in E is in A, and everything in F is in A, so everything that's in at least one of E and F is in A. So E U F is contained in A.

I will use the notation that y is an element of B and x is an element of A.

=====Part I=====

First let's show that

f(E U F) is contained in f(E) U f(F)

Let y be in f(E U F).

By the definition of functions there must be some x in E U F such that

f(x) = y.

By the definition of union however if x is in E U F then x is either in E or F. If x is in E, then f(x) = y is in f(E). Otherwise, x is in F, so f(x) = y is in f(F). Hence, once again by the definition of union y is either in f(E) or f(F) (or both), so we can conclude that y is in f(E) U f(F).

Since this holds for all possible points y in f(E U F), we find that the set

f(E U F) is contained in f(E) U f(F).

=====Part II=====

Next, let's show that

f(E) U f(F) is contained in f(E U F).

If y is in f(E) U f(F) it is in at least one of f(E) or f(F).

If y is in f(E) then there is an x in E such that f(x) = y. Otherwise y is in f(F) and there is an x in F such that f(x) = y.

Now that x must be in E U F, so it also follows that f(x) = y is in f(E U F).

Hence, f(E) U f(F) is contained in f(E U F).

=====Putting it Together=====

We know two things.

(I) f(E U F) contained in F(E) U f(F)

(II) f(E) U f(F) contained in f(E U F)

Whenever we have a set X contained in Y and vice versa, this can only mean that X = Y. Hence, (I) and (II) together give us

f(E U F) = f(E) U f(F)

as desired.

And we're done!

Hope that helps.

Show that if f : A -> B and E,F are subsets of A, then f(E uni F) = f(E) uni f(F) and f(E int F) is a subset

let U be the union operator, ^ intersection

let x be in f(E U F). then there exists y in E U F such that f(y) = x
without loss of generality, y is in E. then x = f(y) is in F(E) subset of f(E) U f(F)
now, let x be in f(E) U f(F). wlog, x is in f(E), so there exists y in E such that f(y) = x. But E subset of E U F, so x is in f(E U F)

let f map reals to reals by f(x) = 2x.
E odd numbers, f evens.
f(E) ^ f(F) = E
f(E^F) = empty set.

Let E denote the set of even int.Prove that w/usual addition, & w/mult. defined by m*n=(1/2)mn, E is a ring?

As you noticed, the only slightly interesting question is the existence of a multiplicative identity element. The multiplicative identity element is 2, because if k is any arbitrary element of the ring, then 2*k = k*2 = (1/2) times 2 times k = k.

I suppose there's some need to prove that E is closed under multiplication:
if m and n are in E, then (1/2)mn is in E because BOTH m and n were even,
hence in ordinary arithmetic mn would be a multiple of 4, so that (1/2)mn is
still a multiple of 2.

Closure under addition is fairly obvious: if m=2j and n=2k in ordinary arithmetic,
then m+n = 2(j+k), and is in E. The zero of addition and the additive inverses
are exactly the same ones that would obtain in the ring of all integers.

All of the usual "ring" properties (associativity, commutative addition, distributive laws, and the existence of additive inverses) are inherited directly from Z (the ring of all integers),
but your teacher may want you to go through them in detail. For instance, the
associativity of multiplication requires that for all m,n,p in E, we would have
(m*n)*p = m*(n*p). Is this true? Of course it is: in the multiplication defined on E,
the result of either of these operations is (1/4)mnp [in the language of ordinary arithmetic].

Prove int(int(S))=int(S)?

Using E as "a member of", it is clear that if x E int(int(S)) then x E int(S), so we need to prove x E int(S)=>x E int(int(S)).

Suppose x E int(S). Then there exists a neighborhood, N such that x E N and N is entirely contained in S. But then N is contained int(int(S)), since all of the points in N are interior points, it being an open set. Hence x E N which is contained in int(int(S))=> x E int(int(S)).

The basic concept is that the interior of an open set is an open set as well and is its own interior.

Hope that helps.