Download Assignment 2, answers.

MATHEMATICS 365 Spring 2017
Assignment #2
Due: Please hand in your own solutions at the start of class on
Friday, Feb.3. Late assignments will not be accepted.
1. Answer the following.
√
√
a) Determine the closures of the intervals (0, 2) and ( 2, 3) in the lower
limit topology and in the topology generated by the basis {[a, b) | a, b ∈
Q, a < b}.[4]
√
Answer. [0, 2) is closed in the lower √
limit topology because it’s complement is open. To see this let, x ∈
/ [0, 2). Then either x < 0, in which
√
−x
case [x, x + ), for √= 2 , is a neighbourhood of x disjoint from [0, √2),
or x ≥ 2, √
in which [ 2, ∞) is a neighbourhood√of x disjoint from [0, 2).
Thus, [0, 2) is a closed subset containing (0, 2).√To show that it√is the
closure, it suffices to observe that the closure
√ of (0, 2) contains (0, 2) of
course, as well as any limit points of (0, 2). But 0 is a limit
√ point, since
any basic open neighbourhood of it (clearly) intersects (0, 2).
√
In
the
‘rational’
lower
limit
topology,
however,
[0,
2) is not closed, because
√
2 is a point√in its complement, any basic open neighbourhood
of which
√
intersects [0, 2), proving
that the complement
We
√
√ of [0, 2) is not open. √
propose instead [0, 2] as the closure of [0, 2). Indeed, the set [0, 2]
is closed in the current topology,
since its complement is open (easy to
√
prove). As
√ the closure of [0, 2) must contain all the limit points
√ of the
set, and 2 is a limit√point, as any√basic open neighbourhood of 2 has the
form
intersects
√ [a, b) with a < 2, and b > 2, and such a neighbourhood
√
[0, 2). Therefore the closure is exactly equal to [0, 2] as claimed.
√
Consider
now
the
interval
(
2, 3). In the lower limit
√ topology, it’s closure is
√
[ 2, 3) by exactly the same argument as for (0, 2). In the ‘rational lower
limit’ topology, 3 is not a limit
√ point, since
√ [3, ∞) is a neighbourhood of
3 which does not intersect
is a limit point, as is easy to
√ ( 2, 3). Also, 2 √
see. So the closure
of ( 2, 3) must contain 2, and so we propose that
√
the closure is [ 2, 3). All that we need to check is that this set is closed;
i.e. that its complement is open, and this follows easily from the above
methods.
b) A subset S ⊂ X of a topological space is said to be dense if its closure S
equals X. For example, Q is dense in R.
Prove that any infinite subset of R is dense in the finite complement topology. [3]
Answer.
Suppose S is an infinite subset of Rc ; we show it is dense. Indeed, the closed sets
in the finite complement topology are exactly the finite sets, together with Rc
itself, whence any closed subset of Rc containing S, being necessarily infinite,
must be Rc . Hence the closure of S, being closed and containing S equals Rc .
So S is dense.
2. Prove that
a) The product of two Hausdorff spaces is Hausdorff. [2]
Answer. If (a, b), (c, d) ∈ X × Y are two distinct points, then either a 6= c
or b 6= d. Suppose a 6= c. Since X is Hausdorff, there exist disjoint open
sets U, V ⊂ X containing a and c respectively. Then U × Y and V × Y are
open subsets of X × Y , containing (a, b) and (c, d) respectively, and they
are disjoint. SImilarly if b 6= d. So X × Y is Hausdorff if X and Y are
Hausdorff.
b) Any subspace of a Hausdorff space is Hausdorff. [2]
Answer.
If S ⊂ X is a subspace, x, y ∈ S, then since X is Hausdorff there exist
disjoint open subsets U, V of X, containing x, y respectively, and then U ∩S
and V ∩ S are open disjoint subsets of S containing x, y respectively.
c) Any order topology is Hausdorff. [3]
Answer. Suppose x 6= y in X, a simply ordered set. Then x < y or y < x,
say x < y. If there exists z such that x < z < y, then U := (−∞z) and
V := (z, +∞) are two disjoint open sets containing x, y respectively.
If on the other hand (x, y) contains no points, then U := (−∞, y) and V :=
(x, +∞) are disjoint (because there is no point between x and y, open, and
containx, y respectively.
3. Let K = { n1 | n = 1, 2, 3, . . .} ⊂ R. Determine the closure of K in each of Rst ,
RK , Rl , and Ru (the ‘upper limit topology’, having basis the sets (a, b], with
a < b), and, finally, in Rf . [5]
Answer. The closure of K in Rst is K ∪ {0} since 0 is clearly a limit point of
K, and since K ∪ {0} is closed.
In RK the closure of K is K since K is closed, since, by definition, it’s complement is open.
In the lower limit topology, 0 is a limit point of K. So the closure of K must
contain 0. On the other hand, K ∪ {0} is closed in the standard topology, and
hence it is closed in the lower limit topology too. It follows that the closure of
K in Rl is K ∪ {0}.
In the upper limit topology, K is closed. The proof follows from the fact that
0 is not a limit point, because (−∞, 0] is a neighbourhood of 0 which doesn’t
meet K. I omit further details.
4. Find a function f : R → R which is continuous at exactly one point. [2]
Answer. Let f (x) = x if x is rational, 0 else. It is routine to check it is
continuous only at 0.
5. Let X be a topological space and A ⊂ X be a dense subset (see Question 1c).)
Prove that if Y is Hausdorff and f and g are two continuous functions X → Y
such that f |A = g|A , then f = g. Give a counter-example if Y is allowed to be
non-Hausdorff. [4]
Answer. Suppose f (x) 6= g(x) for some x. Since Y is Hausdorff, there exist
disjoint open neighbourhoods U, V of f (x), g(x) resp. Then f −1 (U ) and g −1 (V )
are open, and their intersection is open, and contains x, so is non-empty. Since
A is dense there exists a ∈ A in f −1 (U ) ∩ f −1 (V ). By design, f (a) ∈ U , and
g(a) ∈ V , but f (a) = g(a) on A, so f (a) = g(a) ∈ U ∩ V , which contradicts
that U and V disjoint.