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Transcript
```MA4266 Topology
Lecture 17
Wayne Lawton
Department of Mathematics
S17-08-17, 65162749 [email protected]
http://www.math.nus.edu.sg/~matwml/
http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1
Metrization
Definition A topological space ( X , T ) s metrizable if
there exists a metric d on X that gives the topology
Theorem 8.15 The product of a countable number of
metric spaces is metrizable.
Proof

n n 1
(Xn,d )
~
dn
, d n  1 d n : X n  X n  [0, )
are bounded metrics and

~
2
d ( x, y)   n1 (d n ( pn ( x), pn ( y)) / n)
is a metric on


1/ 2
X   n 1 X n that gives the product

topology, see pages 254-255.
T.
Metrization
Thm8.16 Every 2ndcountable regular space is metrizable
Proof Let X be a 2nd countable regular space with basis

n n 1
{B } . Lindelöff’s theorem 6.13 implies X is Lindelöf
hence Theorem 8.9 implies X is normal. Therefore
for each pair (i, j )  N  N with Bi  B j there exists
f : X  [0,1] such that f ( Bi )  0 and
f ( X \ B j )  1. - why? These functions can be
a function

n n 1
F : X  H  2 ( N )
1
1
F
(
x
)

(
f
(
x
),
f
(
x
),...,
by
1
2 2
n f n ( x),...). Then F
indexed { f
}
- why? Define
is an embedding – what exactly does this mean ?
Metrization
Claims
F is one-to-one – why ?
F is continuous – why ?
F : X  F ( X )  H is an open mapping.
Proof pages 257-258 in textbook.
Since
X is homeomorphic to a subspace of H
it is metrizable – why ?
Supplement: The Zarisky Topology
X  F where F is field, for example
F  R, Q, C , Z / pZ where p  N is prime. Let P( X )
denote the ring of polynomials in n variables with
coefficients in F . For A P ( X ) Define the variety of
n
A by V ( A)  {v  F : A(v)  0}. The Zarisky topology
TZar on X is the topology generated by the basis
BZar  { X \ V ( A) : A  P( F )}. See Ex. 4.5.4 125.
n
Definition Let
Examples X  R , X \ V (0)  X \ X   , X \ V (1)  X \   X ,
2
X \ V ( x1 x2 )   4 open quadrants
and
X \ V ( x12  x22  1)  X \ {circle} are open in TZar .
Supplement: The Zarisky Topology
( X , TZar ) is a compact space.
Proof Assume to the contrary that ( X , TZar ) is not
Theorem
compact (we will derive a contradiction) so that there
O with no finite subcover. Then
C  { X \ O : O  O } is a collection of closed sets
exists an open cover
having the finite intersection property (the intersection of
every finite collection of sets in C is nonempty) and
the intersection of all the sets in C is empty. From
these two properties we can find a sequence C j in
with
C
C j  C j 1 (Prove it ! ) so K j   i 1 Ci is a strictly
decreasing sequence of closed sets.
j
Supplement: The Zarisky Topology
j  N define I j  { A  P( X ) : K j  V ( A)}
so I j is the set of all polynomials that vanish on K j .
For each
I j is an ideal in the ring P ( X ). This
means A, B  I j , Q  P( X )  A  B  I j , QA  I j .
Claim 1. Each
Claim 2. I1  I 2  I 3   strictly increases.
Claim 3. I 


j 1
I j is an ideal. ( Prove these claims ! )
Hilbert's basis theorem every ideal in the ring of multivariate
polynomials over a field is finitely generated.
Proof http://en.wikipedia.org/wiki/Hilbert's_basis_theorem
HBT  there exists a basis B  {b1 ,..., bL } for I meaning
I  b1P( X )   bL P( X ). Each bi  I ki so B  I m
with m  max {k1 ,..., kL }.So I m  I a contradiction, why ?.
Assignment 16