Respect to this Prove that a connected space cannot have more than one dispersion points. , I couldn't proof the first item:

Suppose that $p$ is a dispersion point of $X$. $X∖\{p\}$ is totally disconnected, so in particular it is not connected, and there is therefore a non-empty $H⫋X∖\{p\}$ such that $H$ is clopen in $X∖\{p\}$. Let $K=(X∖\{p\})∖H$, the other member of the separation of $X∖ \{p\}$.

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

Ok. Since $A$ and $B$ are a separation of $H\cup\{p\}$, then, by defintion,

  • $A\cap B=\emptyset$.
  • $A=A_1\cap(H\cup\{p\})$ and $B=B_1\cap(H\cup\{p\})$, with $A_1,B_1$ open (or closed) in $X$, and
  • $H\cup\{p\}=A\cup B$\

Then, it is easy to show that $A\cup(K\cup B)=X$ and $A\cap(K\cup B)=\emptyset$. Now, since $p\in B$, then $K\cup B\neq\emptyset$. So, if $A$ and $K\cup B$ are open (or closed) in $X$, there would be a separation of the connected space $X$. Therefore $A=\emptyset$ and thus $H\cup\{p\}$ is connected. But I can't prove the fact that $A$ and $K\cup B$ are open (or closed) in $X$.

Any help or hint is welcome.


$H$ is open in $X\setminus\{p\}$, which is open in $X$, so $H$ is open in $X$. Similarly, $K$ is open in $X$, so $H\cup\{p\}$ is closed in $X$. $A$ is closed in $H\cup\{p\}$ so $A$ is closed in $X$. Finally,

$$A=A_1\cap\big(H\cup\{p\}\big)=A_1\cap H$$

is the intersection of two open sets in $X$ and so is open in $X$. Thus, $A$ is clopen in $X$, and you have your contradiction.

This argument does require that $\{p\}$ be closed in $X$. Since I tend to assume that all spaces under discussion are $T_1$ unless otherwise stated, I probably unconsciously made this assumption when I wrote my answer to the question to which you linked. However, it is possible with a bit more work to give a proof that does not require this assumption.

Lemma. Let $x\in X$ be arbitrary, let $Y=X\setminus\{x\}$, and suppose that $H\subseteq Y$ is clopen in $Y$; then $X\setminus H$ is connected.

Proof. Suppose that $X\setminus H=A\cup B$, where $A\cap B=\varnothing$, $A$ and $B$ are clopen in $X\setminus H$, and $x\in A$. Since $H$ is open in $Y$, there is an open $U$ in $X$ such that $H=U\cap Y=U\setminus\{x\}$; clearly $U=H$ or $U=H\cup\{x\}$. Similarly, there is a closed $C$ in $X$ such that $H=C\cap Y=C\setminus\{x\}$, so $C=H$ or $C=H\cup\{x\}$. Let $W_A$ be an open set in $X$ such that $A=W_A\cap(X\setminus H)$; $x\in W_A$, so

$$W_A\cup H=W_A\cup U\;,$$

which is open in $X$, and its complement

$$X\setminus(W_A\cup H)=(X\setminus W_A)\cap(X\setminus H)=(X\setminus H)\setminus W_A=B$$

is closed in $X$. On the other hand, there is an open $W_B$ in $X$ such that $B=W_B\cap(X\setminus H)$, and since $B\subseteq X\setminus C\subseteq X\setminus H$, we have

$$B=W_B\cap(X\setminus H)\cap W_B\cap(X\setminus H)\cap(X\setminus C)=W_B\cap(X\setminus C)\;,$$

which is open in $X$. Thus, $B$ is a proper clopen subset of $X$, so $B=\varnothing$, and $X\setminus H$ is indeed connected. $\dashv$

Now let $Y=X\setminus\{p\}$, and let $y\in Y$ be arbitrary. I’ll show that $X\setminus\{y\}$ is connected, so that $y$ certainly cannot be a dispersion point of $X$. Suppose that $X\setminus\{y\}=H\cup K$, where $H\cap K=\varnothing$, $H$ and $K$ are clopen in $Y$, and $p\in H$. The lemma ensures that $X\setminus H$ is connected. However, $X\setminus H=K\cup\{y\}$, which is therefore a connected subset of $Y$ containing $y$. Since $Y$ is totally disconnected, we must have $K=\varnothing$, so that $X\setminus\{y\}$ is indeed connected, and $y$ is not a dispersion point of $X$.

  • $\begingroup$ Why the last part? If $\{p\}$ is open in X, which is the contradiction to connectedness of $X$? $\endgroup$ – sinbadh Apr 4 '16 at 22:37
  • $\begingroup$ @sinbadh: Oops! It appears that I was again unconsciously assuming that the space was $T_1$; it’s a deeply ingrained habit. Ignore that part for now; I’ll give it some more thought and either repair it or remove it. $\endgroup$ – Brian M. Scott Apr 4 '16 at 22:44
  • $\begingroup$ @sinbadh: I don’t see a way to remove the assumption that $\{p\}$ is closed from the argument that I sketched before, but I’ve produced a different argument that does not require it. $\endgroup$ – Brian M. Scott Apr 5 '16 at 1:47
  • $\begingroup$ I added "open" to your def'n of $W_A$. Its omission was clearly a typo. $\endgroup$ – DanielWainfleet Apr 5 '16 at 4:00
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    $\begingroup$ @sinbadh: Depends on exactly how you define totally disconnected. If you define it to mean that the components are singletons, then yes, both points of any two-point space are dispersion points. If you add the requirement that the space not be connected, then neither is a dispersion point. But as I said, the whole question is pretty uninteresting for finite spaces. $\endgroup$ – Brian M. Scott Apr 6 '16 at 4:51

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