Download 3-2-2011 – Take-home

FUNDAMENTOS DE TOPOLOGIA E ANÁLISE REAL
1o Semestre 2010/2011
1o Teste (Recuperação) 3-2-2011 – Take-home
1. Let X be a a topological space, ∼ an equivalence relation on X and p : X → X/ ∼
the quotient map. Let R := {(x, y) | x ∼ y} ⊂ X × X.
a) p × p : X × X → (X/ ∼) × (X/ ∼) is continuous.
Answer: p is continuous, by the definition of quotient map. The sets U × V,
with U, V open in X/ ∼ form a basis for the product topology on (X/ ∼) × (X/ ∼)
and (p × p)−1 (U × V) = p−1 (U) × p−1 (V) is open, being the product of open sets.
Hence p × p is continuous.
b) If X/ ∼ is Hausdorff then R is closed in X × X. (Hint: determine A such that
R = (p × p)−1 (A).)
Answer: Recall that X/ ∼ is Hausdorff iff the diagonal ΛX/∼ = {([x], [x]) x ∈ X}
is closed in the product (X/ ∼) × (X/ ∼). Since p × p : X × X → (X/ ∼) × (X/ ∼)
is continuous, and
(p × p)−1 (ΛX/∼ ) = {(x, y) ∈ X × X p(x) = p(y)} = R,
we have that R is closed if ΛX/∼ is closed.
Alternatively: Let (x, y) < R ⇔ [x] , [y], where [x] denotes the equivalence
class of x ∈ X. We check that there is a neighborhood of (x, y) which does not
intersect R, hence (x, y) < R, so R ⊂ R and R is closed1 .
Since X/ ∼ is Hausdorff, there exist disjoint neighborhoods U of [x] and V of [y]
in X/ ∼. Since p is continuous, p−1 (U) and p−1 (V) are open, hence neighborhoods
of x, y in X, respectively, so W := U × V is a neighborhood of (x, y) in X × X.
If (x0 , y0 ) ∈ W then [x0 ] ∈ U and [y0 ] ∈ V. Since U ∩ V = ∅, [x0 ] , [y0 ], that is,
(x0 , y0 ) < R. Hence W ∩ R = ∅.
2. a) Show that there is a topology τ over R such that A ⊂ R is closed if it satisfies at
least one of the following conditions:
1)
0 ∈ A,
2)
A is finite.
Answer: It follows from the definition of topology and from De Morgan’s laws
that τ is a topology iff
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or equivalently, (X × X) \ R is open.
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i) R, ∅ are closed;
ii) ∩i∈J Ai is closed, if Ai closed, arbitrary J;
p
iii) ∪i=1 is closed, if Ai closed, p ∈ Z+ .
As for i), since 0 ∈ R and ∅ is finite, both R, ∅ are closed. Now let Ai be closed,
for any i ∈ J. If Ai is finite for some i ∈ J then ∩i∈J Ai ⊂ Ai is finite, hence closed.
If not, then 0 ∈ Ai , for all i ∈ J, hence 0 ∈ ∩i∈J Ai , which gives that ∩i∈J Ai is
closed and ii) is proved. As for finite unions, if 0 ∈ AI for some i = 1, ..., p, then
p
p
0 ∈ ∪i=1 Ai , hence ∪i=1 Ai is closed. If not, then Ai is finite, for all i = 1, .., p, hence
p
∪i=1 Ai is also finite, hence closed.
NOTE: U ⊂ R open iff 0 < U or U = R \ {x1 , ..., xp }.
b) In (R, τ) any {x}, x , 0, is open but τ is not the discrete topology.
Answer: If x , 0, then 0 ∈ (R \ {x}), hence (R \ {x}) is closed, that is, {x} is open.
On the other hand, {0} is not open, since R \ {0} is not closed: 0 < R \ {0} and
R \ {0} is not finite, so it doesn’t satisfy any of the conditions 1), 2). Hence, τ is
not the discrete topology.
c) Is (R, τ) connected?
Answer: No, since {x}, x , 0 is both open (by b)) and closed (since it is finite),
and not R, ∅. Equivalently: R = {x} ∪ (R \ {x}), x , 0 is a separation. (Can take
any finite A with 0 < A as open and closed.)
d) For infinite A ⊂ R, A = A ∪ {0} under τ. Is (R, τ) separable?
Answer: By definition, A is the intersection of all the closed sets containing A.
Since A is infinite, any X closed with A ⊂ X is infinite, so 0 ∈ X and we have
that 0 ∈ A, so that A ⊃ A ∪ {0}. Since A ∪ {0} is closed and contains A, we have
also A ⊂ A ∪ {0}, and we conclude that A = A ∪ {0}, for infinite A.
(R, τ) separable iff there is a countable dense subset. But if A is finite, it is closed
and A = A , R, and if A is infinite with A = A ∪ {0} = R then A ⊃ R \ 0 is
uncountable, so R is not separable.
e) (R, τ) is regular and not first countable. Is it metrizable?
Answer: (R, τ) is regular iff it is T1 and for any x ∈ R and any closed set A such
that x < A, there are open, disjoint sets U, V such that x ∈ U and A ⊂ V. It is T1,
since any one point set is finite, hence closed. Let now first x , 0. Since {x} is
open, we can take U = {x} and, since {x} is finite, V = R \ {x} is open and A ⊂ V.
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If x = 0 and A closed with 0 < A, then since 0 ∈ R \ A, have that V = A is open,
and 0 ∈ U = R \ A is also open. Hence (R, τ) is regular.
We check that there is no countable basis of neighborhoods at 0. First note that
U is a neighborhood of 0 iff 0 ∈ U and R \ U is finite. Let B = {Bi } be a basis of
neighborhoods at 0, so that R \ Bi is finite, for any i, and for any neighborhood
U of 0, we have U ⊃ Bi for some i, that is, R \ U ⊂ R \ Bi . If B is countable, then
∪R \ Bi is also countable (countable union of finite sets). Take x ∈ R \ ∪(R \ Bi ),
with x , 0. Then U = R \ {x} is a neighborhood of 0 but R \ U = {x} is not
contained in R \ Bi , for any i, which is a contradiction. Hence B is uncountable
and (R, τ) is not first countable.
It is not metrizable, since any metric space is first countable (taking for instance
B(x, 1/n), n ∈ N as a basis of neighborhoods at x).
f) A ⊂ R is compact under τ if and only if it is closed.
Answer: Since (R, τ) is regular, it is in particular Hausdorff, so any compact set
under τ is closed.
Conversely, let A be closed. If A is finite, then it is compact, since any open
covering of a finite set has a finite subcovering. If not, then 0 ∈ A. Let {Uα } be an
open covering of A, and pick α0 such that 0 ∈ Uα0 . Then Uα0 is a neighborhood
of 0, so that R \ Uα0 is finite, and A \ Uα0 is also finite, hence compact. Since {Uα },
α , α0 is an open covering for A \ Uα0 , there is a finite subcover Uα1 , ..., Uαp , and
then
A ⊂ Uα0 ∪ Uα1 ∪ ... ∪ Uαp .
3. Let X be a T1 topological space. Assume that every continuous, bounded, realvalued function on a closed set has a continuous extension to all of X. Show that
X is normal. Is the converse also true?
Answer: X is normal iff it is T1 and for any given closed, disjoint sets A, B, there
are open, disjoint sets U, V such that A ⊂ U, B ⊂ V.
Given that X is T1, we only need to check the second condition. Let then A, B be
closed, disjoint sets. Since A ∩ B = ∅, the function f : A ∪ B → R such that f (A) = 0
and f (B) = 1 is continuous (pasting lemma). It is also bounded and defined on a
closed set, so it has a continuous extension g : X → R, with g(A) = 0 and g(B) = 1.
Define U := g−1 (] − 1/2, 1/2[) and V := g−1 (]1/2, 3/2[). Then U, V are inverse images
of open sets under a continuous function, hence open. Moreover, they are disjoint,
and A ⊂ g−1 (0) ⊂ U, B ⊂ g−1 (1) ⊂ V.
The converse is true, by Tietze’s theorem.
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4. Let Y be a Hausdorff topological space, X a set, and consider the function space
YX with the product topology τp .
a) Show that τp coincides with the topology of pointwise convergence, that is,
fn → f in YX if, and only if, fn (x) → f (x) in Y, for all x ∈ X.
Answer: Can use a general result on the product topology: an → a in Πα∈J Xα
iff πβ (an ) → πβ (a) in Xβ , for all β ∈ J, where πβ : Πα∈J Xβ is the projection. In our
case: J = X and πx ( f ) = f (x).
Alternatively: if fn → f then πx ( fn ) = fn (x) → πx ( f ) = f (x), since πx is continuous. Conversely, let fn , f ∈ YX be such that πx ( fn ) → πx ( f ), for all x ∈ X.
Let U = ΠUx be a neighborhood of f in the product topology, with Ux = Y,
for x , x1 , ..., xp and Uxi is a neighborhood of f (xi ) in Y. Since fn (xi ) → f (xi ),
for each i = 1, ..., p, have that there is Ni such that if n > Ni then fn (xi ) ∈ Uxi .
Let N = max Ni . Then n > N gives that fn (xi ) ∈ Uxi , for any i = 1, ..., p, that is,
fn (x) ∈ Ux , for any x ∈ X, that is, fn ∈ U. Hence, fn → f .
b) F ⊂ YX is compact
if and only if F is closed and, for any x ∈ X, the set
f (x) | f ∈ F has compact closure in Y. (Hint: write F = Πx∈X πx (F ), with
πx : YX → Y the projection map.)
Answer: Note that πx (F ) = f (x) | f ∈ F . Let F ⊂ YX be compact. Since YX is
Hausdorff, being the product of Hausdorff spaces, F is closed. Moreover, since
the projection πx is continuous, for any x ∈ X, have that πx (F ) is compact in Y.
Since Y is Hausdorff, it is also closed, so πx (F ) = πx (F ) is compact.
Conversely: if F is closed, then F = F = Πx∈X πx (F ) = Πx∈X πx (F ) is compact,
by Tychonoff’s theorem, since πx (F ) is compact.
c) Let τ be a topology on YX finer than τp . If YX is compact under τ, then τ = τp .
Answer: The identity map (YX , τ) → (YX , τp ) is continuous, since τ ⊃ τp , and
any continuous bijection from a compact space to a Hausdorff space is a closed
map, therefore a homeomorphism. Hence τ = τp .
d) Let (Y, d) be a compact metric space, X be a topological space. What can you say
about compactness of YX in the pointwise convergence, uniform, and compact
convergence topologies?
Answer: In the pointwise convergence topology: it follows from b) 2 that YX is
compact iff Y is compact. In the uniform and compact convergence topologies,
it may not be compact.
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or Tychonoff’s theorem together with πx continuous
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Let τc , τu denote the compact convergence and uniform topologies, respectively.
Then we have τp ⊂ τc ⊂ τu . Hence, it follows from c) that YX is compact in τc ,
respectively, in τu , iff τc = τp , respectively, τu = τp .
It follows from the definition of τc that τc = τp iff the only compact sets in X
are finite (this holds if X is discrete). On the other hand, τu = τp iff X is finite (
τu = τc iff X is compact).
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