Is the topology τ a Hausdorff Space in R²?

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Homework Statement


Let τ be a topology consisting of ∅, \Re^{2} and complements of finite unions of points and lines. Is this a hausdorrf space?


Homework Equations


Just definition of Hausdorrf space
given a topological space X, the diagonal X\timesX is closed iff X is hausdorrf.


The Attempt at a Solution



My first thoughts were that every set in the space is an open set in the topology. Therefore, take the diagonal {(x,x) | x\inX} and the complement of that. That will be open. therefore \Re is hausdorrf in the space. And the product of hausdorrf space is hausdorrf so the topology is hausdorrf?
 
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First of all, it's spelled Hausdorff :)


happysauce said:
My first thoughts were that every set in the space is an open set in the topology.

Well, that ain't right. Think more carefully.

Start from the definition of Hausdorff spaces. So let x and y be points in X, then try to find disjoint open sets that contain x and y.
 
clamtrox said:
First of all, it's spelled Hausdorff :)




Well, that ain't right. Think more carefully.

Start from the definition of Hausdorff spaces. So let x and y be points in X, then try to find disjoint open sets that contain x and y.

I think I figured it out. It isn't hausdorrf. Suppose it was hausdorrf. Then given two points x1 and x2 i can find neighborhoods that are disjoint. Namely U and V. If x1 is in U then X2 is in V\subset R^2 - U, which is a countable set of lines and points and does not exist in the topology and is a contradiction.

I read about a similar proof showing the finite complement topology wasn't hausdorrf and realized this is the same idea with any space. I guess I could claim that in general, any space R^n with the complement topology isn't hausdorrf because disjoint unions would imply one of them is not in the topology!
 
happysauce said:
I think I figured it out. It isn't hausdorrf.

Hausdorff after Felix Hausdorff, a German mathematician of the first half of the 20th century. Technical terms named after people are always capitalized.

Hausdorff studied at the University of Leipzig, obtaining his Ph.D. in 1891. He taught mathematics in Leipzig until 1910, when he became professor of mathematics at the University of Bonn. He was professor at the University of Greifswald from 1913 to 1921. He then returned to Bonn. When the Nazis came to power, Hausdorff, who was Jewish, felt that as a respected university professor he would be spared from persecution. However, his abstract mathematics was denounced as "Jewish", useless, and "un-German"[citation needed] and he lost his position in 1935. Though he could no longer publish in Germany, Hausdorff continued to be an active research mathematician, publishing in the Polish journal Fundamenta Mathematicae.

After Kristallnacht in 1938 as persecution of Jews escalated, Hausdorff became more and more isolated. He wrote to George Pólya requesting a research fellowship in the United States, but these efforts came to nothing.[1]

Finally, in 1942 when he could no longer avoid being sent to a concentration camp, Hausdorff committed suicide together with his wife, Charlotte Goldschmidt Hausdorff, and sister-in-law, Edith Goldschmidt Pappenheim,[2] on 26 January. They are buried in Bonn, Germany.


http://en.wikipedia.org/wiki/Felix_Hausdorff
 
happysauce said:
I think I figured it out. It isn't hausdorrf. Suppose it was hausdorrf. Then given two points x1 and x2 i can find neighborhoods that are disjoint. Namely U and V. If x1 is in U then X2 is in V\subset R^2 - U, which is a countable set of lines and points and does not exist in the topology and is a contradiction.

I read about a similar proof showing the finite complement topology wasn't hausdorrf and realized this is the same idea with any space. I guess I could claim that in general, any space R^n with the complement topology isn't hausdorrf because disjoint unions would imply one of them is not in the topology!

That doesn't sound right. What about the two sets R^2-{x1} and R^2-{x2}?
 
Dick said:
That doesn't sound right. What about the two sets R^2-{x1} and R^2-{x2}?

{x1} and {x2} aren't neighborhoods though right?
Also {x1} and {x2} are not in the topology or open sets either.
 
Last edited:
happysauce said:
{x1} and {x2} aren't neighborhoods though right?
Also {x1} and {x2} are not in the topology or open sets either.

R^2-{x1} is the complement of a point. Doesn't that mean it's in the topology?
 
Dick said:
R^2-{x1} is the complement of a point. Doesn't that mean it's in the topology?

You're confusing T1-spaces and Hausdorff spaces.
 
clamtrox said:
You're confusing T1-spaces and Hausdorff spaces.

Indeed I am. Thanks for the correction!
 
  • #10
Thanks for the help guys! :P
 
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