I don't have the time to tackle all of this at once, but here are a few answers:
Notation: I'm using _{...} and ^{...} to indicate subscripts and
superscripts respectively.
1. (=>) Suppose X is Hausdorff. Let x, y be distinct elements of X.
Then there exist disjoint open sets A_{x,y}, B_{x,y} containing x, y
respectively.
Now A_{x,y} x B_{x,y} is a basis element in the product topology. Let
P be the union over all distinct x, y in X of A_{x,y} x B_{x,y}. Then
P is open in X x X. We show that P is the complement of D and
therefore D is closed.
Now clearly for any distinct x, y in X (x, y) is in P by definition,
so P contains the complement of D. Also, since each A_{x,y} and
B_{x,y} are distinct, no element A_{x,y} x B_{x,y} contains any
element of D, and so their union P also contains no elements of D.
Hence P is the complement of D. So D is the complement of an open set,
i.e. D is closed.
(<=): Suppose D is closed in X x X. Note that the open sets of X x X
are simply the unions of sets of the form A x B where A, B are open in
X (this follows from the definition of the product topology and the
fact that the intersection of any two such sets is another such set,
so intersections generate no extra sets). The complement of D is open,
and can therefore be written as some union of sets {A_i x B_i : i in
I} with each A_i and B_i open in X; since A_i x B_i is in the
complement of D it follows that A_i and B_i are disjoint for all i.
Now if x, y are distinct elements of X, (x, y) is in the complement of
D and therefore must be in A_i x B_i for some particular i. Then A_i
and B_i are disjoint open sets containing x and y respectively, so X
is Hausdorff.
2. Interior of A = union of all open sets contained in A
Closure of A = intersection of all closed sets containing A =
complement of the interior of the complement of A.
Now int1(A) is the union of all subsets X of A which are elements
of T1. Since T1 is a subset of T2, all such X are also elements of T2,
so int1(A) is a subset of int2(A). Similarly clos2(A) is a subset of
clos1(A) - either by an analogous argument or by relating it to the
interior of the complement of A as above.
Examples: take R with T1 the usual topology, T2 the discrete topology
(every subset is open). Let A = [0, 1] and B = (0, 1).
Then int1(A) = (0, 1), int2(A) = [0, 1]
clos1(B) = [0, 1], clos2(B) = (0, 1).
3. (a) Note that intersections of base elements only give us other
base elements, so open sets are unions of open intervals and (sets of)
rational numbers. I don't really see how we can make the description
much simpler. You could say that an open set is a set such that for
every irrational element there is an open interval containing that
element which is contained in the set (roughly speaking "irrational
elements only occur in open intervals"), but that's not simpler to my
mind.
(b) Yes - we can use the same open sets as normal, since this is a
finer topology than the usual one. Specifically, if x and y are
distinct real numbers then take open intervals (x-d, x+d) and (y-d,
y+d) where d = |x-y|/2.
(c) Yes. Let A be a closed set (basically a normal closed set, but
with holes at arbitrary (perhaps all!) rational points) and let x be a
point not in A. If x is rational, (-oo, x) u (x, oo) is an open set
containing A and {x} is an open set containing x.
If x is irrational, suppose that every neighbourhood of x contains a
point of A. Then we can construct a sequence (a_n) of distinct points
of A with limit x (in R). (Let d_0 = 1, a_n be a point of A within d_n
of x, and d_{n+1} = d(a_n, x)/2. Then a_0 is within 1 of x and each
successive a_n is at most half the distance from x of the preceding
one, so a_n is within 1/(2^n) of x.) Now A is a subset of some set B
that is closed in R with the usual topology, with B\A a subset of Q.
So the sequence (a_n) is also contained in B; since B is closed with
the usual topology, the limit x must also be in B. But then B\A
contains x, a contradiction. So there must be some neighbourhood of x
that is disjoint from A. Let this neighbourhood be U. Then U must
contain an open interval containing x (see part (a)), say (y, z). But
any open interval containing x contains a smaller interval containing
x, say (w, v). Then (w, v) is an open neighbourhood of x and (-oo, w)
u (v, oo) is an open neighbourhood of A.
I can't help feeling that there ought to be a simpler argument. But I
haven't found one.
(d) No - for instance (-oo, 0) and [0, oo) are disjoint open sets
whose union is R.
And skipping a few...
8. The metric space approach is useful here. I will show A is complete
and totally bounded.
A is complete: Let (x_n) be a Cauchy sequence in A, and suppose it has
no limit in A. It certainly has a limit in R, say c. Define f: A->R by
f(x) = 1/(x-c). Then f is continuous on A, but {f(x_n)} is unbounded
(since (x_n) converges to c in R, we can get arbitrarily close to c
and hence arbitrarily large values of f). This is a contradiction, so
(x_n) must have a limit in A and A is complete.
A is totally bounded: The identity function from A to R is bounded,
hence A must be contained in some interval [m, M]. Let e > 0. Now [m,
M] is compact in R, so we can cover it with finitely many open balls
B_1, ..., B_n of radius e/3. Discard the ones that are disjoint from
A, and renumber the remaining ones as B_1, ..., B_m. Let c_k be a
point of C_k = A n B_k for each k. Then the open ball B(c_k, e)
contains all of C_k (since the maximum distance of any two points in
B_k is 2e/3) and hence their union contains A. So A is totally
bounded. Since A is complete and totally bounded, it is compact.
I'll have a look at questions 4-7 if I have time. |
