Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A topological space $X$ is called relative extremely disconnected if it has a base $B$ (for open subsets) such that disjoint elements in $B$ have disjoint closure. Does it exist an infinite Hausdorff space $X$ which is not relative extremely disconnected?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Hello dear Ali. I think the answer is yes. consider the closed unit interval $I=[0,1]$, and define the set $K$ as follows:$$K=I\times I -(0,1)\times (0)$$

roughly speaking eliminate the interval $(0,1)$ from the bottom of the unit square.

Now we are to define the base of each point of $K$.

If $(x,y)\neq (0,0) , (1,0)$ define the neighborhoods to be as in the usual Euclidean topology.

If $(x,y)=(0,0)$ define the base to be all the sets $[0,\frac{1}{2})\times (0,\epsilon)$, where $\epsilon>0$.

If $(x,y)=(1,0)$ define the base to be all the sets $(\frac{1}{2},1]\times(0,\delta)$, where $\delta>0$.

It is obvious to see that this new space is Hausdorff. But it is not relatively extremely disconnected. To see this consider any neighborhoods of $(0,0)$ and $(1,0)$. it is intuitive to see that the closure of these neighborhoods intersect each other in some point at the edge $x=\frac{1}{2}$.

share|improve this answer
    
Hi Alireza Your answer is excelent and so if you would we can do a common work about this space. –  Ali Nov 24 '12 at 20:08
    
This is wrong. According to the original post, a space is relatively extremely disconnected if there is any base $B$ such that disjoint elements of $B$ have disjoint closure. We can modify your base such that this is the case, say by replacing $[0,1/2)$ with $[0,1/3)$ and $(1/2,1]$ with $(2/3,1]$. –  Alex Becker Nov 26 '12 at 2:28
    
Dear Alex. Please check it more precisely. You could not change the intervals in x-axis to them. please look at the basis of $(0,0)$ and $(1,0)$. each of neighborhoods of $(0,0)$ geometrically should contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. also each of neighborhoods of $(1,0)$ contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. so you could not change them in your favor. –  Ali Reza Nov 26 '12 at 6:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.