Prove that a connected space cannot have more than one dispersion points. Prove that a connected space cannot have more than one dispersion points. 

I have no idea at all how can I able to tackle the problem.can anyone help me please.
Also can I get some examples of topological spaces having dispersion point
 A: This Wikipedia entry has two examples of a space with a dispersion point. Another example is the Alexandroff extension $\Bbb Q^*$ of $\Bbb Q$, the space of rational numbers with the usual topology: $\Bbb Q$ is totally disconnected, but every open nbhd of the point in $\Bbb Q^*\setminus\Bbb Q$ is dense in $\Bbb Q^*$, which is therefore connected.
Suppose that $p$ is a dispersion point of $X$. $X\setminus\{p\}$ is totally disconnected, so in particular it is not connected, and there is therefore a non-empty $H\subsetneqq X\setminus\{p\}$ such that $H$ is clopen in $X\setminus\{p\}$. Let $K=\big(X\setminus\{p\}\big)\setminus H$, the other member of the separation of $X\setminus\{p\}$.


*

*Show that $H\cup\{p\}$ is connected. HINT: If $A$ and $B$ are a separation of $H\cup\{p\}$ with $p\in B$, what can you say about the sets $A$ and $K\cup B$?

*Conclude that if $q\in K$, then $q$ cannot be a dispersion point of $X$.

*Interchange the rôles of $H$ and $K$ to show that no point of $H$ can be a dispersion point of $X$ either, and hence $p$ is the only dispersion point of $X$.
