most recent 30 from http://mathoverflow.net 2013-05-25T13:52:20Z http://mathoverflow.net/feeds/question/114314 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114314/relative-extremely-disconnected-space Ali 2012-11-24T05:55:26Z 2012-11-27T20:32:39Z

<p>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 <em>it</em> exist <em>an</em> infinite Hausdorff space $X$ which is not relative extremely disconnected?</p>

http://mathoverflow.net/questions/114314/relative-extremely-disconnected-space/114347#114347 AliReza Olfati 2012-11-24T16:49:05Z 2012-11-24T20:36:33Z

<p>Hello dear Ali. I think the answer is yes. consider the closed unit interval $I=[0,1]$, and define the set $K$ as follows:$$K=I\times I -(0,1)\times (0)$$</p> <p>roughly speaking eliminate the interval $(0,1)$ from the bottom of the unit square.</p> <p>Now we are to define the base of each point of $K$. </p> <blockquote> <p>If $(x,y)\neq (0,0) , (1,0)$ define the neighborhoods to be as in the usual Euclidean topology.</p> <p>If $(x,y)=(0,0)$ define the base to be all the sets $[0,\frac{1}{2})\times (0,\epsilon)$, where $\epsilon>0$.</p> <p>If $(x,y)=(1,0)$ define the base to be all the sets $(\frac{1}{2},1]\times(0,\delta)$, where $\delta>0$.</p> </blockquote> <p>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 $x=\frac{1}{2}$. </p>