Download NOTES WEEK 13 DAY 1 Unassigned exercise: Let X and Y be

NOTES WEEK 13 DAY 1
SCOT ADAMS
Unassigned exercise: Let X and Y be topological spaces, let s P X and
t P Y . Let W be a nbd of ps, tq in X ˆ Y . Show that there exists nbds
U of s in X and V of t in Y such that U ˆ V Ď W .
Unassigned exercise: Let X and Y be topological spaces. Define p :
X ˆ Y Ñ X and q : X ˆ Y Ñ Y by ppx, yq “ x and qpx, yq “ y. Show
that p : X ˆ Y Ñ X and q : X ˆ Y Ñ Y are both continuous.
Assigned HW#46.
REMARK 0.1. Let X and Y be topological spaces. Let a P pX ˆ Y qN .
Define p : X ˆ Y Ñ X and q : X ˆ Y Ñ Y by ppx, yq “ x and
qpx, yq “ y. Assume that ppa‚ q is convergent in X and that qpa‚ q is
convergent in Y . Then a‚ is convergent in X ˆ Y .
Proof. This is immediate from HW#46.
DEFINITION 0.2. Let X be a topological space and let a P X N . Then
a is convergent in X means: Dz P X s.t. a‚ Ñ z in X. Also, a is
subconvergent in X means: Dsubsequence b‚ of a‚ , Dz P X s.t. b‚ Ñ z
in X.
After Definition 0.2, a topological space X is sequentially compact
iff every sequence in X is subconvergent.
Unassigned exercise: Let X be a topological space, let S Ď X and let
a P S N . Then:
( [ a‚ is convergent in S ] ñ [ a‚ is convergent in X ) ] and
( [ a‚ is subconvergent in S ] ñ [ a‚ is subconvergent in X ] ).
Let X :“ R2 . Let D be a closed disk in R2 and let p P BX D. We
pictured S :“ Dztpu. We pictured a sequence in a P S N such that
a‚ Ñ p in X. We noted that, because p R S, it follows that a‚ is
Date: November 29, 2016
Printout date: November 30, 2016.
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not convergent in S. So, in the next unassigned exercise (below), it is
important to assume that S is closed in X.
Unassigned exercise: Let X be a topological space, let S Ď X and let
a P S N . Assume that S is closed in X. Then:
( [ a‚ is convergent in X ] ñ [ a‚ is convergent in S ] ) aand
( [ a‚ is subconvergent in X ] ñ [ a‚ is subconvergent in S ] ).
DEFINITION 0.3. Let pX, dq be a metric space and let a P X N .
Then a‚ is Cauchy in pX, dq means: @ε ą 0, DK P N s.t. @i, j P N,
p i, j ě K q
ñ
p dpai , aj q ă ε q.
We assigned HW#47, HW#48 and HW#49.
LEMMA 0.4. Let Z be a set. Let X be a topological space. Let a P Z N
and let f : Z Ñ X. Assume that X is sequentially compact. Then there
is a subsequence b‚ of a‚ such that f pb‚ q is convergent in X.
Proof. Let α‚ :“ f pa‚ q. Then α P X N , so, as X is sequentially compact,
choose a subsequence β‚ of α‚ such that β‚ is convergent in X. Choose
a strictly increasing k P NN such that β‚ “ αk‚ . Let b‚ :“ ak‚ . We wish
to show that f pb‚ q is convergent in X. Since β‚ is convergent in X, it
suffices to show that f pb‚ q “ β‚ . We wish to show, for all j P N, that
pf pb‚ qqj “ βj . Let j P N be given. We wish to show that pf pb‚ qqj “ βj .
Since b‚ “ ak‚ , it follows that bj “ akj . Since f pa‚ q “ α‚ , it follows
that f pakj q “ αkj . Since αk‚ “ β‚ , it follows that αkj “ βj . Then
pf pb‚ qqj “ f pbj q “ f pakj q “ αkj “ βj , as desired.
LEMMA 0.5. Let X be a topological space, let a, b P X N . Assume
that a‚ is convergent in X and that b‚ is a subsequence of a‚ . Then b‚
is convergent in X.
Proof. This follows from Lemma 0.1 Week 12 Day 1.
LEMMA 0.6. Let X be a set. let a, b, c P X N . Assume that b‚ is
a subsequence of a‚ and that c‚ is a subsequence of b‚ . Then c‚ is a
subsequence of a‚ .
Proof. Unassigned exercise.
THEOREM 0.7. Let X and Y be sequentially compact topological
spaces. Then X ˆ Y is sequentially compact.
NOTES WEEK 13 DAY 1
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Proof. We wish to show: For any a P pX ˆ Y qN , there exists a subsequence c‚ of a‚ such that c‚ is convergent in X ˆ Y . Let a P pX ˆ Y qN
be given. We wish to show that there exists a subsequence c‚ of a‚
asuch that c‚ is convergent in X ˆ Y .
Define p : X ˆ Y Ñ X and q : X ˆ Y Ñ Y by ppx, yq “ x and
qpx, yq “ y. By Lemma 0.4, choose a subsequence b‚ of a‚ such that
ppb‚ q is convergent in X. By Lemma 0.4, choose a subsequence c‚ of b‚
such that qpc‚ q is convergent in Y . By Lemma 0.6, c‚ is a subsequence
of a‚ . We wish to show that c‚ is convergent in X ˆ Y .
Since c‚ is a subsequence of b‚ , it follows, from Remark 0.6 Week 12
Day 1, that ppc‚ q is a subsequence of ppb‚ q. Then, by Lemma 0.5, since
ppb‚ q is convergent in X, we see that ppc‚ q is convergent in X. Since
ppc‚ q is convergent in X and qpc‚ q is convergent in Y , it follows, from
Remark 0.1, that c‚ is convergent in X ˆ Y , as desired.
We observed that, while R2 ‰ R ˆ R, the topological spaces R2 and
R ˆ R are homeomorphic. So, while R2 and R ˆ R are technically
different from one another, they “look the same” to a topologist. A
similar argument shows that, for any n P N, for any S1 , . . . , Sn`1 Ď R,
while pS1 ˆ ¨ ¨ ¨ ˆ Sn q ˆ Sn`1 Ď Rn ˆ R and S1 ˆ ¨ ¨ ¨ ˆ Sn ˆ Sn`1 Ď Rn`1
are not equal, it is nevertheless true that pS1 ˆ ¨ ¨ ¨ ˆ Sn q ˆ Sn`1 and
S1 ˆ¨ ¨ ¨ˆSn ˆSn`1 are homeomorphic, and the natural homeomorphism
to use is the (anonymous) function:
ppx1 , . . . , xn q, yq ÞÑ px1 , . . . , xn , yq
:
pS1 ˆ ¨ ¨ ¨ ˆ Sn q ˆ Sn`1 Ñ S1 ˆ ¨ ¨ ¨ ˆ Sn ˆ Sn`1 .
THEOREM 0.8. For any n P N, for any C1 , . . . , Cn Ď R, we have:
p C1 , . . . , Cn are all sequentially compact q
ñ
p C1 ˆ ¨ ¨ ¨ ˆ Cn is sequentially compact q.
Proof. For all n P N let Pn be the proposition
‚ for any C1 , . . . , Cn Ď R, we have:
p C1 , . . . , Cn are all sequentially compact q
p C1 ˆ ¨ ¨ ¨ ˆ Cn is sequentially compact q.
We wish to show: @n P N, Pn . For all C1 Ď R, we have
p C1 is sequentially compact q
p C1 is sequentially compact q.
ñ
ñ
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Then P1 . So, by the Prinicple of Mathematical Induction, it suffices to
show, for all n P N, that Pn ñ Pn`1 . Let n P N be given. We wish to
show that Pn ñ Pn`1 . Assume Pn . We wish to show Pn`1 . That is,
we wish to show
‚ for any C1 , . . . , Cn , Cn`1 Ď R, we have:
p C1 , . . . , Cn , Cn`1 are all sequentially compact q
ñ
p C1 ˆ ¨ ¨ ¨ ˆ Cn ˆ Cn`1 is sequentially compact q.
Let C1 , . . . , Cn , Cn`1 Ď R be given. We wish to show:
p C1 , . . . , Cn , Cn`1 are all sequentially compact q
ñ
p C1 ˆ ¨ ¨ ¨ ˆ Cn ˆ Cn`1 is sequentially compact q.
Assume that C1 , . . . , Cn , Cn`1 are all sequentially compact. We wish to
show that C1 ˆ¨ ¨ ¨ˆCn ˆCn`1 is sequentially compact. By Pn , we know
that C1 ˆ¨ ¨ ¨ˆCn is sequentially compact. So, since Cn`1 is sequentially
compact, it follows, from Theorem 0.7, that pC1 ˆ ¨ ¨ ¨ ˆ Cn q ˆ Cn`1 is
sequentially compact. So, since
pC1 ˆ ¨ ¨ ¨ ˆ Cn q ˆ Cn`1
and
C1 ˆ ¨ ¨ ¨ ˆ Cn ˆ Cn`1
are homeomorphic, we conclude that C1 ˆ¨ ¨ ¨ˆCn ˆCn`1 is sequentially
compact, as desired.