# Connected sets and separations

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 $$X$$.

• Why the last part? If $\{p\}$ is open in X, which is the contradiction to connectedness of $X$? Commented Apr 4, 2016 at 22:37
• @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. Commented Apr 4, 2016 at 22:44
• @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. Commented Apr 5, 2016 at 1:47
• I added "open" to your def'n of $W_A$. Its omission was clearly a typo. Commented Apr 5, 2016 at 4:00
• @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. Commented Apr 6, 2016 at 4:51