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Let $p$ a cut point of a connected space $(X, \mathcal{T})$ and suppose $C$ and $D$ form a separation of $X-\{p\}$, i.e $$X-\{p\}=C \cup D$$ Prove that $C \cup \{p\}$ is conected.

I have this idea: Since $p$ is a cut point then $X-\{ p \}$ is not connected, and since $C$ and $D$ form a separation of $X- \{p \}$ then $X-\{p \} = C \cup D$, and also $C \cup D$ is not connected. And also $$C \cap \overline{D}= \overline{C} \cap D= \emptyset$$ I need an idea to continue please

Alejandra Benítez
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$\newcommand{\cl}{\operatorname{cl}}$HINT: Suppose that $C\cup\{p\}$ is not connected; then there are sets $A$ and $B$ that form a separation of $C\cup\{p\}$, and we may assume that $p\in B$. We know that $C\cap\cl D=\varnothing$, so $\cl D\subseteq D\cup\{p\}\subseteq D\cup B$. Use this to get a contradiction by showing that $A$ and $D\cup B$ are a separation of $C$.

Note added 15 October 2021: I am assuming that the OP really did mean that $C$ and $D$ form a separation of $X\setminus\{p\}$. The OP’s paraphrase of that is incorrect even partition was actually intended.

Brian M. Scott
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  • It is true that $C\cap\cl D=\varnothing$, but that requires a proof. In fact, one can show that either both $C,D$ are open in $X$ or closed in $X$, but this is not completely trivial. – Paul Frost Oct 06 '20 at 11:17
  • @PaulFrost: There is a reason that the answer is labelled HINT. And since $C$ and $D$ form a separation of $X\setminus{p}$, it’s trivial that $C\cap\operatorname{cl}D=\varnothing$. – Brian M. Scott Oct 06 '20 at 16:09
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    Of course a hint is not an elaborated answer, and once we know that $C \cap \text{cl} D = \varnothing$, then everything is straightforward. My point was that $C \cap \text{cl} D = \varnothing$ is not trivial (unless we accept without proof that ${p}$ is open or closed in $X$). – Paul Frost Oct 06 '20 at 16:47
  • @PaulFrost: Of course it’s trivial that $C\cap\operatorname{cl}D=\varnothing$: by definition $C$ and $D$ form a separation of $X\setminus{p}$ iff $C\cup D=X\setminus{p}$ and $C\cap\operatorname{cl}D=\varnothing=D\cap\operatorname{cl}C$. The OP’s paraphrase is incorrect, and I assumed that separation was in fact meant. – Brian M. Scott Oct 15 '21 at 17:32
  • Okay, now I understand your point. You understand a separation of a subspace $Y \subset X$ as a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point in $X$ of the other (as mentioned in the answer to https://mathoverflow.net/q/30509). I understand a separation as on p.148 in Munkres' book or in https://proofwiki.org/wiki/Definition:Separation_(Topology). Probably there is no standard definition of a separation. In your interpretation $C \cap \text{cl} D = \emptyset$ is trivial (part of the definition), in my interpretation it needs a proof. – Paul Frost Oct 15 '21 at 23:01
  • By the way, the above question is a special case of Munkres' exercise 12 on p.152. – Paul Frost Oct 15 '21 at 23:05
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A partition of a space $Y$ is a pair $(A,B)$ of disjoint nonempty open subsets of $Y$ such that $Y = A \cup B$. Note that then $A, B$ are clopen (= open and closed) in $Y$.

Let us first prove the following:

Let $X$ be a connected topological space, let $x$ be a cut point of $X$ and let $(C,D)$ be a partition of $X \setminus \{x\}$. Then $\{x\}$ is either open or closed in $X$. Moreover, if $\{x\}$ is open, then $C$ and $D$ are closed in $X$, and if $\{x\}$ is closed, then $C$ and $D$ are open in $X$.

Proof. Since $C$ is clopen in $X \setminus \{x\}$, there is an open subset $V$ of $X$ such that $C = V \cap (X \setminus \{x\}) = V \setminus \{x\}$ and there is a closed subset $F$ of $X$ such that $C = F \cap (X \setminus \{x\}) = F\setminus \{x\}$. Thus $V \setminus \{x\} = F \setminus \{x\}$. But $V = F$ is impossible because then $V$ would be a nonempty clopen subset of $X$ whose complement $X \setminus V = (X \setminus \{x\}) \setminus (V \setminus \{x\}) = (X \setminus \{x\}) \setminus C = D$ is a nonempty which contradicts the fact that $X$ is connected. Hence either $x \in V$ and $x \notin F$ or $x \in F$ and $x \notin V$. In the first case we have $\{x\} = V \setminus F$ which is open in $X$, in the second case we have $\{x\} = F \setminus V$ which is closed in $X$. The "moreover" part is obvious since $C, D$ are clopen in $X \setminus \{x\}$, thus in the first case closed in the closed subspace $X \setminus \{x\}$ and in the second case open in the open subspace $X \setminus \{x\}$. Note that $\{x\}$ cannot be both open and closed - otherwise $(\{x\}, X \setminus \{x\})$ would be a partition of $X$.

Now let us prove that $C' = C \cup \{x\}$ is connected. The same is of course true for $D' = D \cup \{x\}$.

Let $\mathbf 2 = \{1,2\}$ with the discrete topology. It is well-known that a space $Z$ is connected iff all continuous maps $Z \to \mathbf 2$ are constant.

If $\{x\}$ is open, then $C,D$ are closed in $X$ and $C',D'$ are open as their complements. Similarly, if $\{x\}$ is closed, then $C',D'$ are closed in $X$.

Now let $f : C' \to \mathbf 2$ be continuous. Define $F : X \to \mathbf 2, F(x) = f(x)$ for $x \in C'$ and $F(x) = f(p)$ for $x \in D'$. Note that this is well-defined because on $C' \cap D' = \{x\}$ both parts agree.

By definition, the restrictions $F \mid_{C'}$ and $F \mid_{D'}$ are continuous. Since either both $C', D'$ are open or are closed, $F$ is continuous. Since $X$ is connected, $F$ is constant. Thus also $f = F \mid_{C'}$ is constant.

Paul Frost
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  • Actually my question has been linked with this question and your answer. Your proof seems to be an overkill( I don't mean any offence or disrespect ) and something which experts in subject can outline. But as a beginner I tried to do it following way:Attempt: I proved A and B are connected using contradiction. Now , For A$\bigcup ${p} to be connected I need to only prove that $Closure (A) \bigcup ${p} $\neq \phi$. I assumed let p doesnot belongs to closure of A and trying to get some contradiction but I am unable to do so." In case you have some spare time, and you are.. –  Oct 16 '21 at 08:16
  • interested can you please outline a proof on the basis of assuming that the converse holds for the purpose of getting a contradiction. Can you please help with that? –  Oct 16 '21 at 08:18
  • @Avenger You should precisely define what you understand by a separation of $X \setminus {p}$; see the comments after Brian M. Scott's answer. If you use "separation" in the sense that $A, B$ are separated in $X$, then his answer tells you everything you need. If you understand it in the way I did (I called it partition), then I encourage you to write an answer proving that $A$ and $B$ are connected. I will then check what is missing to completely answer the question. – Paul Frost Oct 16 '21 at 08:51