# 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

• i think quotient space formed by identifying irrationals to a point p in real line will be a space having p as a dispersion point but i am not sure. Commented May 4, 2013 at 6:17
• anybody has confirmed answer for that.please give me the answer. Commented May 4, 2013 at 10:43

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 $X$.

• And just to be complete, the result that a connected space cannot have more than one dispersion point requires that the space has at least three points (because a connected space with two points trivially has two dispersion points). I am really interested to know, in your experience, what's the usual definition of dispersion point (the usual one ... , or with some added cardinality assumption)? Commented Nov 22, 2023 at 2:43
• @PatrickR: The Wikipedia definition incorporates a requirement of at least three points, but I’ve seen the definition given without it. I really have no idea what’s most common. I’ve checked all of the textbooks that I have here, and only Dugundji’s has an index entry for the term. He requires the space to be Hausdorff, though $T_1$ is good enough: one really just needs to exclude the Sierpiński space. Commented Nov 22, 2023 at 7:02
• Wikipedia is not really an authoritative source. Many papers define dispersion point without extra assumptions. I would say you don't even need $T_1$ for the definition to make sense. For example the particular point topology (of which the Sierpinski space is a special case) has a dispersion point. (I was asking you because someone was wondering if the definition in pi-base needs to be tweaked somehow) Commented Nov 22, 2023 at 7:27
• Also fyi, the OP's result is a classical thing, due to Kline (1922): A theorem concerning connected point sets Fundamenta Mathematicae 3 (1922), 238-239. I have also looked at the classic Knaster & Kuratowski (1921) where they introduce the notion of biconnectedness and their fan, etc. But it's surprising how "sloppy" all these early authors were sometimes. One has to infer things, like the sets they are talking about being nonempty, etc. And of course, $X$ with at least three points for the result above. Commented Nov 22, 2023 at 7:33
• Thanks for all your comments. Always appreciated! Commented Nov 22, 2023 at 19:36