Difference between revisions of "07081300/Class notes for Tuesday, April 1"
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Hence the compliment of an n1 sphere in <math>S^n</math> has two connected components. This is the Jordan Curve Theorem.  Hence the compliment of an n1 sphere in <math>S^n</math> has two connected components. This is the Jordan Curve Theorem.  
+  
+  
+  ===Second Hour===  
+  
+  
+  '''Excersize'''  
+  
+  Consider the embedding of a line in <math>\mathbb{R}^3</math> which wraps arround somewhat like a helix only each loop makes a link with each previous loop and in addition the helical object gets smaller and converges to two points at the end.  
+  
+  The question is, should we put a little circle L about a part of the helical object can we make this L the boundary of a disk embedded in <math>R^3</math>.  
+  
+  Sketch of Answer: Consider the bubble which has L as the boundary and goes all the way arround the object at one end. This is not an embedding since there are two intersection points with the helical object. Cut two small holes from the bubble at each of these points. This is still not quite it since the boundary also has these two small holes. Hence connect the two holes via a tubular neighbourhood going along the helical object and we have the result.  
+  
+  The real excersize here is to figure out how the Mayer Vietoris sequence actually determines this particular surface.  
+  
+  
+  '''Proposition'''  
+  
+  An open path connected set in <math>\mathbb{R}^n</math> is clopen connected.  
+  
+  ''Proof'' This is easy, just using general topology  
+  
+  
+  Hence, for open sets in <math>\mathbb{R}^n</math> path components are the same as clopen components  
+  
+  
+  
+  '''Theorem'''  
+  
+  "Invariance of Domain" or "Openness in <math>\mathbb{R}^n</math> is intrinsic"  
+  
+  
+  If <math>V\subset\mathbb{R}^n</math> is homeomorphic to an open set in <math>\mathbb{R}^n</math> then it is an open set in <math>\mathbb{R}^n</math>  
+  
+  
+  Here are two examples that illustrate why this might not be obvious  
+  
+  1) Replacing open with dense would not be true, for instance, the rationals in [0,1] are not dence in <math>\mathbb{R}</math> but they ARE homeomorphic to the rationals which ARE dense in R.  
+  
+  2) Replacing open with closed is also not true, for instance, <math>\mathbb{R}^n</math> is closed in <math>\mathbb{R}^n</math>  
+  
+  but it is also homeomorphic to the open ball which is not closed.  
+  
+  
+  ''Proof''  
+  
+  Let <math>\phi:U\rightarrow B</math>  
+  
+  Consider a ball B in U, with boundary the circle S and we get a disk <math>D=B\cup S</math>  
+  
+  We have that <math>\phi(S)</math> devides <math>\mathbb{R}^n_V</math> into two connected components. Call them <math>A_1</math> and <math>A_2</math>.  
+  
+  We have both <math>\phi(B)</math> and <math>\mathbb{R}^m  \phi(D)</math> but we need to show these ARE the two connected components. We first note that both are in fact connected.  
+  
+  <math>(R^n \phi(D))\cup\phi(B) = \mathbb{R}^n\phi(S) = A_1\cup A_2</math>  
+  
+  Hence <math>\phi(B)</math> is one of the components and so it is open. ''Q.E.D''  
+  
+  
+  
+  '''Theorem''' BorsukUlam  
+  
+  If <math>g:S^n\rightarrow R^n</math> is continuous then <math>\exists x\in S^n</math> such that g(x) = g(x)  
+  
+  
+  '''Corrollary''' The Ham Sandwich Theorem (or Salad Bowl Theorem, according to Dr. Bar Natan)  
+  
+  Loosely this says that if we have n objects in a bounded domain of <math>\mathbb{R}^n</math> such as items in a salad or sandwhich then we can find an n1 dimensional hyperplane such that the hyperplane precisely cuts each item in half.  
+  
+  Formally, If <math>\mu_1,\cdots,\mu_n</math> are densities in <math>\mathbb{R}^n</math> with bounded support then <math>\exists</math> a hyperplane <math>H\subset\mathbb{R}^n</math> deviding <math>\mathbb{R}^n</math> into <math>H_+\cup H_</math> such that <math>\forall j</math>, <math>\mu_j(H_+) = \mu_j(H_)</math>  
+  
+  
+  ''Proof:''  
+  
+  Consider the set of sided hyperplanes in <math>\mathbb{R}^n</math> which equals <math>S^{n1}_v\times R_t</math> where the first component determines the normal vector v and the second component determines how far away this plane is front the origin in the direction of v.  
+  
+  Define <math>\bar{g}:S^{n1}\times\mathbb{R}\rightarrow\mathbb{R}^n</math> via  
+  
+  <math>(v,t)\mapsto (\mu_1(H_+),\cdots, \mu_n(H_))</math>  
+  
+  Now, if t>>0, clearly <math>\bar{g}(v,t) = 0</math> and if t<<0 then <math>\bar{g}(v,t) = (\mu_1(\mathbb{R}^n),\cdots,\mu_n(\mathbb{R}^n))</math>  
+  
+  Hence we have a cylinder mapping into <math>\mathbb{R}^n</math> such that it is constant along the top and bottom of the cylindar. Hence, we can pinch the top and bottom of the cylinder. We thus get a map g from <math>S^n</math> to <math>\mathbb{R}^n</math> and, applying BorsukUlam, deduce there is a point such that g(x) = g(x) and so <math>\bar{g}(v,t) = \bar{g}(v,t)</math> for some point (v,t). Hence <math>\mu_i(H_+)=\mu_i(H_)</math> for all i  
+  
+  ''Q.E.D'' 
Revision as of 12:17, 1 April 2008

Typed Notes
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.
First Hour
Theorem
If is an embbeding then
Proof:
By induction on k.
For k=0 this is easy, is a single point and  a point is just
Now suppose we know the theorem is true for k1, let be an embedding. Write the cube as where and
So,
where both sets in the union are open. Recall for sets where A and B are open we have the Mayer Vietoris Sequence.
Note: It is usefull to "open the black box" and think about the Mayer Vietoris Sequence from the veiwpoint of singular homology, ie using maps of simplexes and the like. This was briefly sketched in class.
We get a sequence
The first and last terms in this sequence (at least the small part of it we have written) vanish by induction hypothesis.
Note: Technically we need to check at Mayer Vietoris sequence also works for reduced homology.
We thus get the isomorphism:
Assume
As the above isormorphism is induced by the inclusion maps at the chain level, we get that either in or in
Now repeat, find a sequence of intervals such that
1)
2) in
Pictorially, we cut C in half, find which half is non zero, then cut this half in half and find the one the is non zero in and again cut it in half, etc...
But, in by induction. Hence, such that .
But, = { union of images of simplexes in U} is compact, so, for some j. But this contradicts assumption 2)
Q.E.D
We now prove an analogous theorem for spheres opposed to cubes
Theorem
For an embedding then if p = nk1 and zero otherwise.
Intuitively,
Recall we had seen this explicitly for when we wrote as the union of two tori (an inflated circle)
Proof:
As in the previous proof, we use Mayer Vietoris. Let
Then,
The first and last term written vanish by induction hypothesis and so have an isomorphism
Hence, it is enough to prove the theorem for k=0. Well in this case we have
Since , this is for p = n1 and 0 otherwise
Q.E.D
Take k=n1 so
Hence the compliment of an n1 sphere in has two connected components. This is the Jordan Curve Theorem.
Second Hour
Excersize
Consider the embedding of a line in which wraps arround somewhat like a helix only each loop makes a link with each previous loop and in addition the helical object gets smaller and converges to two points at the end.
The question is, should we put a little circle L about a part of the helical object can we make this L the boundary of a disk embedded in .
Sketch of Answer: Consider the bubble which has L as the boundary and goes all the way arround the object at one end. This is not an embedding since there are two intersection points with the helical object. Cut two small holes from the bubble at each of these points. This is still not quite it since the boundary also has these two small holes. Hence connect the two holes via a tubular neighbourhood going along the helical object and we have the result.
The real excersize here is to figure out how the Mayer Vietoris sequence actually determines this particular surface.
Proposition
An open path connected set in is clopen connected.
Proof This is easy, just using general topology
Hence, for open sets in path components are the same as clopen components
Theorem
"Invariance of Domain" or "Openness in is intrinsic"
If is homeomorphic to an open set in then it is an open set in
Here are two examples that illustrate why this might not be obvious
1) Replacing open with dense would not be true, for instance, the rationals in [0,1] are not dence in but they ARE homeomorphic to the rationals which ARE dense in R.
2) Replacing open with closed is also not true, for instance, is closed in
but it is also homeomorphic to the open ball which is not closed.
Proof
Let
Consider a ball B in U, with boundary the circle S and we get a disk
We have that devides into two connected components. Call them and .
We have both and but we need to show these ARE the two connected components. We first note that both are in fact connected.
Hence is one of the components and so it is open. Q.E.D
Theorem BorsukUlam
If is continuous then such that g(x) = g(x)
Corrollary The Ham Sandwich Theorem (or Salad Bowl Theorem, according to Dr. Bar Natan)
Loosely this says that if we have n objects in a bounded domain of such as items in a salad or sandwhich then we can find an n1 dimensional hyperplane such that the hyperplane precisely cuts each item in half.
Formally, If are densities in with bounded support then a hyperplane deviding into such that ,
Proof:
Consider the set of sided hyperplanes in which equals where the first component determines the normal vector v and the second component determines how far away this plane is front the origin in the direction of v.
Define via
Now, if t>>0, clearly and if t<<0 then
Hence we have a cylinder mapping into such that it is constant along the top and bottom of the cylindar. Hence, we can pinch the top and bottom of the cylinder. We thus get a map g from to and, applying BorsukUlam, deduce there is a point such that g(x) = g(x) and so for some point (v,t). Hence for all i
Q.E.D