10-327/Classnotes for Thursday September 30 - Revision history

Xwbdsb at 17:31, 4 October 2010

2010-10-04T17:31:40Z

← Older revision		Revision as of 13:31, 4 October 2010
Line 20:		Line 20:

	[[User:D\|D]] 12:48, 2 October 2010 (EDT) finite Hausdorff space: The only topology satisfying Hausdorff for a finite set A is the discrete topology; for each distinct point x and y in A, {x} and {y} are disjoint. Also, every finite point set in A is closed. We can check that using the discrete metric on A (d(x,y) = 1 if x =/= y, d(x,y) = 0 if x=y), a sequence of points in A can converge to only one point at most. Also note that Hausdorff condition is stronger than T1.		[[User:D\|D]] 12:48, 2 October 2010 (EDT) finite Hausdorff space: The only topology satisfying Hausdorff for a finite set A is the discrete topology; for each distinct point x and y in A, {x} and {y} are disjoint. Also, every finite point set in A is closed. We can check that using the discrete metric on A (d(x,y) = 1 if x =/= y, d(x,y) = 0 if x=y), a sequence of points in A can converge to only one point at most. Also note that Hausdorff condition is stronger than T1.

			Discrete topology implies Hausdorrf condition is obvious. The converse we can prove by contradiction so that its much clearer. Suppose the space X is Hausdorff but it does not have discrete topology. So some finite subset A would not be open. A complement is a finite subset of X would not be closed. That cannot happen because Hausdorff condition implies T1 so that any finite union of singleton set should still be closed. Contradiction here. -Kai [[User:Xwbdsb\|Xwbdsb]] 13:31, 4 October 2010 (EDT)

D at 16:48, 2 October 2010

2010-10-02T16:48:04Z

← Older revision		Revision as of 12:48, 2 October 2010
Line 16:		Line 16:

	[http://katlas.math.toronto.edu/drorbn/images/f/f3/10-327-lec06p05.jpg Lecture 6 page 5]		[http://katlas.math.toronto.edu/drorbn/images/f/f3/10-327-lec06p05.jpg Lecture 6 page 5]



			[[User:D\|D]] 12:48, 2 October 2010 (EDT) finite Hausdorff space: The only topology satisfying Hausdorff for a finite set A is the discrete topology; for each distinct point x and y in A, {x} and {y} are disjoint. Also, every finite point set in A is closed. We can check that using the discrete metric on A (d(x,y) = 1 if x =/= y, d(x,y) = 0 if x=y), a sequence of points in A can converge to only one point at most. Also note that Hausdorff condition is stronger than T1.

Drorbn at 21:05, 30 September 2010

2010-09-30T21:05:09Z

← Older revision		Revision as of 17:05, 30 September 2010
Line 1:		Line 1:
	{{10-327/Navigation}}		{{10-327/Navigation}}

			Some blackboard shots are at {{BBS Link\|10_327-100930-143624.jpg}}.

	{{Template:10-327:Dror/Students Divider}}		{{Template:10-327:Dror/Students Divider}}

Jdw at 20:18, 30 September 2010

2010-09-30T20:18:27Z

New page

{{10-327/Navigation}}

{{Template:10-327:Dror/Students Divider}}

Here are some lecture notes..

[http://katlas.math.toronto.edu/drorbn/images/b/b7/10-327-lec06p01.jpg Lecture 6 page 1]

[http://katlas.math.toronto.edu/drorbn/images/6/65/10-327-lec06p02.jpg Lecture 6 page 2]

[http://katlas.math.toronto.edu/drorbn/images/8/8d/10-327-lec06p03.jpg Lecture 6 page 3]

[http://katlas.math.toronto.edu/drorbn/images/5/5b/10-327-lec06p04.jpg Lecture 6 page 4]

[http://katlas.math.toronto.edu/drorbn/images/f/f3/10-327-lec06p05.jpg Lecture 6 page 5]