Download Assignment 2 SOLUTION MATH 6540 (1) Show that a subset K of a

Assignment 2 SOLUTION
MATH 6540
(1) Show that a subset K of a topological space is closed and discrete (i.e., is a closed subset and is
discrete in the subspace topology) if and only if every subset of K is closed in X.
SOLUTION: Suppose that K is closed and discrete. Fix A ⊆ K. Since every subset of a discrete
space is closed in that space, it follows that A is closed in K. But since K is closed in X, it
follows that A is closed in X.
Conversely, suppose that every subset of K is closed in X. Then K is closed in X (being a
subset of itself ). Fix any subset A of K. A is closed in X, so A is also closed in K. Therefore
every subset of K is closed in K. So the subspace topology on K is the discrete topology.
a
(2) Consider [0, 1]2 with the order topology determined by the lexicographic order on [0, 1]2 . Let
A = [0, 1] × {1/2}. Find A.
SOLUTION: A = A ∪ (0, 1] × {0} ∪ [0, 1) × {1}. To see this, let us first fix some notation: We
will use hx, yi to denote ordered pairs and the parentheses (, ), [, ] for interval notation. Since the
complement of A ∪ (0, 1] × {0} ∪ [0, 1) × {1} is equal to the union of the open intervals:
[h0, 0i, h0, 1/2i) ∪ (h1, 1/2i, h1, 1i] ∪
[
{(hx, 0i, hx, 1/2i) ∪ (hx, 1/2i, hx, 1i) : x ∈ (0, 1)}
it follows that the set A ∪ (0, 1] × {0} ∪ [0, 1) × {1} is closed.
So it suffices to show that every point hx, ii ∈ (0, 1] × {0} ∪ [0, 1) × {1} is a limit point of
A. First consider hx, 0i for x > 0. Since x > 0, any open interval around hx, 0i is of the form
J = (ha, bi, hc, di) where a < x. If we fix y such that a < y < x we see that hy, 1/2i lies between
ha, bi and hx, 0i hence A ∩ J 6= ∅. So for any x ∈ (0, 1] we have that hx, 0i is a limit point of A.
By a similar argument, any point of the form hx, 1i for x < 1 is a limit point of A. Indeed,
since x < 1, any open interval around hx, 1i is of the form J = (ha, bi, hc, di) where x < c. If we
fix y such that x < y < c we see that hy, 1/2i lies between hx, 1i and hc, di hence A ∩ J 6= ∅. So
for any x ∈ [0, 1) we have that hx, 1i is a limit point of A.
a
(3) Suppose that X and Y are spaces and that Y is Hausdorff. Suppose that f : X → Y is
continuous. Prove that the graph of f is a closed subset of X × Y . Conversely, prove that if the
graph of the identity function from Y :→ Y is closed, then Y is Hausdorff.
SOLUTION: To see that the graph of f is closed, fix (x, y) ∈ X × Y not on the graph of f . So
y 6= f (x). By the Hausdorf property of Y , there are disjoint open sets U and V such that y ∈ U
and f (x) ∈ V . By continuity of f we have that f −1 (V ) is an open neighbourhood of x. Therefore
(x, y) ∈ W = f −1 (V ) × U . It now suffices to show that W is disjoint from the graph of f since
this would imply that (x, y) is not a limit point of the graph and hence the graph of f is closed.
To see that W is disjoint from the graph, consider a point on the graph (a, f (a)) and suppose
that a ∈ f −1 (V ). So then f (a) ∈ V . But since V ∩ U = ∅ it follows that f (a) 6∈ U . Therefore
(a, f (a)) 6∈ W .
For the converse, suppose that ∆ = {(y, y) : y ∈ Y } is closed in Y × Y and fix x 6= y in Y .
So (x, y) 6∈ ∆. So there are open sets U and V such that
(i) (x, y) ∈ U × V and
(ii) U × V ∩ ∆ = ∅.
Therefore it follows from (i) that x ∈ U and y ∈ V and we now claim that U ∩ V = ∅. Indeed,
if not, then any z ∈ U ∩ V would correspond to a point (z, z) ∈ ∆ ∩ U × V contradicting (ii). a
2
(4) We observed in class that the Sorgenfrey line has the peculiar property of having a base consisting
of clopen sets. Which of the following spaces also have this property?
(a) Q with the Euclidean subspace topology it inherits from the real line.
SOLUTION: Yes, Q is zero-dimensional. To see this, not that if x, y are irrational, then
(x, y) ∩ Q is open in Q and also (x, y) ∩ Q = [x, y] ∩ Q, hence is also closed in Q.
So it suffices to show that
B = {(x, y) ∩ Q : x, y ∈ R \ Q}
is a base for the topology on Q. But this follows easily from the density of the irrationals
in the reals. Indeed, suppose that U is open in Q and that q ∈ U . Then by definition of
the subspace topology there is a W open in R such that U = W ∩ Q. And since q ∈ W ,
there is an interval (a, b) around q contained in W . Choosing irrationals x and y such that
a < x < q < y < b we have that q ∈ (x, y) ∩ Q ⊆ U as required.
a
(b) [0, 1] × {0, 1} with the lexicographic order topology
SOLUTION: Yes, this space is zero-dimensional. We saw in class that we had a basis
consisting of the sets {h0, 0i} and {h1, 1i} (since these points are isolated) and sets of the
form
(hx, 0i, hy, 1i)
where 0 ≤ x < y ≤ 1. So these intervals are open, but they are also closed since
(hx, 0i, hy, 1i) = [hx, 1i, hy, 0i]
(c) [0, 1] with the usual topology
SOLUTION: This is a rather deeper result and I really didn’t expect you to solve this unless you
had either seen it before, or looked it up. We will do this in class soon.
(5) Prove directly that TFAE for metric space (X, d) and (Y, m), a function f : X → Y and a point
x0 ∈ X
(a) For every A ⊆ X, if x0 ∈ A then f (x0 ) ∈ f (A)
(b) For every > 0 there is a δ > 0 such that m(f (x0 ), f (x)) < whenever d(x0 , x) < δ.
SOLUTION: (a)⇒(b): Assume (a) and fix > 0. The first case is that x0 is an isolated
point of X. If so, there is a δ > 0 such that B(x0 , δ) = {x0 }. So if d(x, x0 ) < δ then x = x0
so it follows that f (x0 ) = f (x) and so m(f (x0 ), f (x)) = 0 < .
Now assume that x0 is not isolated and by way of contradiction, assume that there is no
δ satisfying the conclusion of (b). Now construct a sequence (yn ) converging to x0 by
choosing, for each n, a yn so that d(x0 , yn ) < 1/n but m(f (x0 ), f (yn )) > (here we use
that the conclusion of (b) fails for δ = 1/n). Then Letting A = {yn : n ∈ N} we have that
x0 ∈ A but f (x0 ) 6∈ f (A) contradicting (a).
(b)⇒(a): Assume (b) and suppose that x0 ∈ A. To see that f (x0 ) ∈ f (A) fix > 0. It
suffices to show that B(f (x0 ), )∩A 6= ∅. By (b) there is a δ > 0 such that m(f (x0 ), f (x)) < whenever d(x0 , x) < δ. Since x0 ∈ A, we may fix y ∈ A such that d(x0 , y) < δ. Hence
m(f (x0 ), f (y)) < and since f (y) ∈ f (A) we have B(f (x0 ), ) ∩ A 6= ∅ as required.
a
(6) Prove that ω1 is not metrizable. Hint: If it were, then for every α there would be an n such that
the ball of radius 1/n about α would be contained in (β, α] for some β < α.
3
SOLUTION: It is easiest to first note that any infinite subset of ω1 has a limit point. To see
this, let α be minimal in ω1 such that B ∩ [0, α) is infinite. Then for any β < α, B ∩ [0, β) is
finite (by minimality of α), so B ∩ (β, α] is infinite. So α is a limit point of B.
Next, suppose that ω1 is metrizable with its topology generated by the metric d. Since successor elements of ω1 are isolated, for each such successor element x there is an nx such that
B(x, 1/nx ) = {x}. Since there are uncountably many successor elements and N is countable,
there is a single n ∈ N such that B = {x : nx = n} is infinite. Notice that any two distinct elements of B are of distance ≥ 1/n apart. Now, let α be a limit point of B. Now
fix < 1/(2n). So there are x0 , x1 ∈ B ∩ B(α, ). But now we have a contradiction, for
1/n ≤ d(x0 , x1 ) ≤ d(α, x0 ) + d(α, x1 ) < + < 1/(2n) + 1/(2n) = 1/n.
a