# Relative extremely disconnected space

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?

-

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 $x=\frac{1}{2}$.

-
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