Download Homework 2 - Connor O`Dell

Homework 2
Connor O’Dell
(3A1). If F is the collection of all closed bounded subset of Rstd together
with R itself, then F is the family of closed sets for a topology on R strictly
weaker than the usual topology.
Proof. The intersection of any number of closed and bounded sets is again
closed and bound. The first fact is clear and the last fact is true as the
intersection must be contained in any one of the bounded sets in the intersection. Similarly, any finite union of closed and bounded sets is again
closed and bounded. R is in the family by assumption, and ∅ is both closed
and bounded. Thus F is a family of closed sets and defines a topology on R
by Theorem 3.4. The open sets of this topology are a subset of the topology
on Rstd , as theSare all the complements of closed sets in Rstd .
Note that n∈N (n, n + 1) cannot be in this new topology, as its complement N is not bounded, but is open in Rstd .
(3A4). Show that the collection of radially open sets is a topology for R2 .
Compare this with Rstd .
First note that ∅ is open vacuously, and R is radially open as it contains
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all line segments. Suppose Ui is radially opend for i ∈ I. Let U = i∈I Ui .
Let x ∈ U . Then x ∈ Uj for some j ∈ I. Thus there is an open line segment
in each direction about x contained in Ui . These are also contained in U , so
U is radially open.
Suppose V1 and V2 are radially open. If y ∈ V1 ∩ V2 , there exist line
segments in each direction about y in both V1 and V2 . For each direction
about y, the shortest of these two line segments must be in both V1 and V2 .
Thus V1 ∩ V2 is open. Induction shows that any finite intersection of open
sets is open.
Let τ be the set of radially open sets of R2 . We can see that τ is finer
than Rstd . This is true as all open balls are also radially open, which implies
that all open sets of Rstd are also open in τ . But τ has more open sets than
Rstd as we can take the unit ball at zero and append both axes and still
have an open set.
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MATH 5610
Connor O’Dell
(3A5). If A ⊂ X and τ is any topology for X, then {U ∪ (V ∩ A) | U, V ∈ τ }
is a topology for X.
Proof. Let σ = {U ∪ (V ∩ A) | U, V ∈ τ }. Suppose Si ∈ σ for i ∈ I. Then
[
[
S=
Si = (Ui ∪ (Vi ∩ A))
for some Ui , Vi ∈ τ
i∈I
i∈I
=
[
Ui
[
∪
i∈I
=
[
(Vi ∩ A)
i∈I
Ui
[
∪
i∈I
Vi
∩
A
i∈I
which is in σ as τ must be closed under arbitrary unions.
If S1 , S2 ∈ σ,
S1 ∩ S1 = (U1 ∪ (V1 ∩ A) ∩ (U2 ∪ (V2 ∩ A)
= ((U1 ∪ (V1 ∩ A)) ∩ U2 ) ∪ ((U1 ∪ (V1 ∩ A)) ∩ (V2 ∩ A))
= (U1 ∩ U2 ) ∪ (V1 ∩ U2 ∩ A) ∪ (U1 ∩ V2 ∩ A) ∪ (V1 ∩ V2 ∩ A)
= (U1 ∩ U2 ) ∪ [(V1 ∩ U2 ) ∪ (U1 ∩ V2 ) ∪ (V1 ∩ V2 )] ∩ A
which is in σ as τ is closed under finite intersections. We also see that
∅ = ∅ ∪ (∅ ∩ A) and X = X ∪ (X ∩ A), so ∅, X ∈ σ.
(3D1). The complement of a regularly open set is regularly closed and vice
versa.
Proof. Suppose G is regularly open. Then in particular, Gc is closed and
int(C c ) ⊆ Gc . Suppose F is a closed set containing int(Gc ). We show that
Gc ⊆ F , and hence Gc is the smallest closed set containing int(Gc ) and so
equal to int(Gc ) ⊆ Gc .
If Gc * F , ∃x ∈ Gc such that x ∈
/ F . Since F is closed (and F c is open),
there exists an open set U such that x ∈ U ⊆ F c ⊆ G. thus U ∩ Gc = ∅, a
contradiction of the fact that Gc is closed. Hence Gc ⊆ F .
Now suppose G is regularly closed. Then Gc is open and Gc ⊆ int(Gc ),
as Gc ⊆ Gc . We show that any open set contained in Gc is also contained
in Gc , and hence int(Gc ) ⊆ Gc .
Suppose V ⊆ Gc is open. Recall that Gc = Gc ∪ ∂Gc . If V * Gc ,
∃x ∈ ∂Gc = ∂G ⊆ G. The last containment follows from the fact that G
is closed. Since G is regularly closed, x must be in the interior of G or the
closure of the interior of G. The first case is clearly impossible. In the second
case, there must be some y ∈ V such that y ∈ int(G). Then there exists
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MATH 5610
Connor O’Dell
some open set A ⊆ G with y ∈ A. Now y ∈ A ∩ V ⊆ V , and A ∩ V is open.
But V * Gc as y cannot be in Gc . Therefore we must have V ⊆ Gc .
(3D2). There are open sets in R which are not regularly open.
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Proof. U = n∈Z (n, n + 1) is open in Rstd and int(U ) = int(R) = R which
is clearly not equal to U .
(3D3). If A is any subset of a topological space, then int(cl(A)) is regularly
open.
Proof.
int(cl(int(cl(A)))) = int(cl(int(A ∪ ∂A)))
= int(cl((A ∪ ∂A) ∩ (∂(A ∪ ∂A)c )))
= int(cl((A ∪ ∂A) ∩ (∂A)c ))
= int(cl((A ∪ ∂A) ∩ (int(A) ∪ int(Ac ))))
= int(cl(int(A)))
= int(cl(A)).
(3D4). The intersection but not necessarily the union of two regularly open
sets is regularly open.
Proof. We see that (−∞, 0) and (0, ∞) are both regularly open in Rstd but
the interior of the closure of their union is R. So the union of two regularly
open sets is not necessarily regularly open.
Let W = int(U ∩ V ). If U and V are regularly open, then U ∩ V is open
and contained in U ∩ V , so U ∩ V ⊆ W .
For the reverse containment, we first note the X is open and contained
in U and V . So we also seet that W ⊆ int(U ) = U and W ⊆ int(V ) = V .
Thus W ⊆ U ∩ V .
(3F2). If ρ generates the topology on a metrizable space X and, for each
λ ∈ Λ, Cλ isSa closed set in X such that ρ(Cλ1 , Cλ2 ) ≥ for all λ1 , λ2 , and
> 0, then λ∈Λ Cλ is closed.
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Proof. Let C = λ∈Λ Cλ . Let x ∈ C, and fix > 0. Then B(x, ) ∩ C 6= ∅.
But any element of B(x, ) can only belong to one Cλ , so x ∈ Cλ = Cλ ⊆ C.
Hence C = C, and C is closed.
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