Theorem 23.3 of Munkres’ Topology

Question

The union of a collection of connected subspace of $X$ that have a point in common is connected.

The following attempt is inspired by https://math.stackexchange.com/a/4384059/861687 post.

My attempt: let $f: \bigcup_{\alpha \in J} A_\alpha \to \{ 0,1\}$ be continuous map. By theorem 18.2, map $f|_{A_\alpha}:A_\alpha \to \{0,1\}$ defined by $f|_{A_\alpha}(x)=f(x),\forall x\in A_{\alpha}$ is continuous for each $\alpha \in J$. Since $A_\alpha$ is connected, we have $f|_{A_\alpha}(A_\alpha)=f(A_\alpha)=\{ i_\alpha \}$, $i_\alpha \in \{ 0,1\}$ and $i_\alpha$ depends on each $\alpha$. Since $\bigcap_{\alpha \in J} A_\alpha \neq \phi$, $\exists p\in \bigcap A_\alpha$. So $p\in A_\alpha, \forall \alpha \in J$. $p\in \bigcup A_\alpha$. Suppose $f(p)=i$, for some $i\in \{0,1\}$. Then $f(p)=i=i_\alpha ,\forall \alpha \in J$. Thus $i=i_\alpha, \forall \alpha \in J$. Let $x\in \bigcup A_\alpha$. Then $x\in A_\beta$, for some $\beta \in J$. So $f(x)=i_\beta =i$. Thus $f(x)=i, \forall x\in \bigcup A_\alpha$. Hence $f$ Is constant and $\bigcup_{\alpha \in J} A_\alpha$ is connected. Is this proof correct?

In this proof, we don’t make use of lemma 23.2 and exercise 1 of section 16 like Munkres’ proof.

Edit: Proof of equivalent definition of connected space: https://courses-archive.maths.ox.ac.uk/node/view_material/50743, proposition 1.77, page number 17. In lecture notes, it’s given $(1)\Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. One can also use $(1)\Leftrightarrow (2) \Leftrightarrow (3)$. Munkres have already proved $(1) \Leftrightarrow (2)$ and $(2) \Rightarrow (3)$ is given in lecture notes. Claim: $(3)\Rightarrow (2)$. Proof: Assume towards contradiction, i.e. $\exists P,Q\in \mathcal{T}_X$ such that $P,Q\neq \phi$, $P\cap Q=\phi$ and $P\cup Q=X$. Let $f: X\to \{0,1\}$ map defined by $f(P)=0$ and $f(Q)=1$. $f$ is a well defined map. It is easy to check $f^{-1}(\{0\})=P$ and $f^{-1}(\{1\})=Q$. Note $\mathcal{T}_Y =\{ \phi ,\{0\},\{1\}, \{0,1\} \}$. $\forall V\in \mathcal{T}_Y, f^{-1}(V)\in \mathcal{T}_X$. So $f$ is continuous but not constant. Which contradicts our initial assumption.

Answer 1

This is not a full verification of your proof as your post mentions other results and doesn't complete the argument: $f$ constant implies what, exactly? Other than that however, your argument seems correct to me.

I leave this here as a demonstration of how it can be simpler, with no other theorems referenced.

$\newcommand{\C}{\mathcal{C}}\newcommand{\O}{\mathcal{O}}$A simple proof I produced myself for the same exercise (with weaker assumptions - it could well be the case that the entire intersection is empty):

Let $\{\C_\lambda:\lambda\in\Lambda\}$ be a nonempty collection of nonempty connected subsets of a topological space $X$, with each member having a common point with every other member, pairwise. Then $\bigcup_{\lambda\in\Lambda}\C_\lambda$ is a connected subspace of $X$.

Proof:

The case $|\Lambda|=1$ is trivial, so take at least $2$ elements therein.

Suppose $\O_1,\O_2$ disconnect $\bigcup_{\lambda\in\Lambda}\C_{\lambda}$. There must exist at least one $\lambda\in\Lambda$ for which $\O_1\cap\C_{\lambda}\neq\emptyset$. If $\O_2\cap\C_{\lambda}\neq\emptyset$, then $\O_1,\O_2$ would disconnect $\C_{\lambda}$ which is impossible, so take $\O_2\cap\C_\lambda=\emptyset$ and thus $\C_\lambda\subseteq\O_1$.

There exists $\lambda'\in\Lambda$ with $\lambda'\neq\lambda$ such that $\C_\lambda\cap\C_{\lambda'}\neq\emptyset$ and $\O_2\cap\C_{\lambda'}\neq\emptyset$ by assumption and by $|\Lambda|\gt1$. But as $\C_\lambda\subseteq\O_1$, we have $\C_{\lambda'}\cap\C_\lambda\neq\emptyset$ implies $\O_1\cap\C_{\lambda'}\neq\emptyset$. Then $\O_1,\O_2$ disconnect $\C_{\lambda'}$ which is again impossible.

Therefore no disconnection of nonempty disjoint open sets exists of $\bigcup_{\lambda\in\Lambda}\C_\lambda\,\blacksquare$

Answer 2

Yes, that proof is correct, using the alternate characterisation using functions into $\{0,1\}$ we discussed earlier.

You could also start with $p$ and note that for each $\alpha$ we must have by constantness that $f\restriction_{A_\alpha} \equiv f(p)$ from the start. It then immediately follows that $f \equiv f(p)$ and we're done.