Strengthening of a Kuratowski's theorem about Connected Sets

It is asserted in this article on "Connective Spaces" that in topological spaces, if $A \subseteq B$ are both connected and $C$ is maximally connected in $B - A$, then $B - C$ is connected (page 4, proposition (a)). I have not been able to prove this, so could someone post a proof of it? The same page 4 of the article lists two propositions (b) and (c) which can be proven from (a); could someone also show how each of them are proven?

In addition, I am trying to prove that if $A \subseteq B$ are both connected and $C$ is maximally connected in $B - A$, then $A \cup C$ is connected. Does anyone know if this is true? I've been stuck on this for very long...

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Try this one: math.stackexchange.com/questions/157175/… –  Sigur Aug 26 '12 at 18:05

In topological language, (a) says that if $A$ and $B$ are connected, $A\subseteq B$, and $C$ is a connected component of $B\setminus A$, then $B\setminus C$ is connected.

Proof: Let $X=B\setminus C$, and suppose that $X$ is not connected; then there are disjoint sets $G$ and $H$ such that $X=G\cup H$, and $G$ and $H$ are clopen in $X$. $A\subseteq X$, and $A$ is connected, so $A\subseteq G$ or $A\subseteq H$; without loss of generality $A\subseteq G$. Let $U=G\cup C$; then $U\cap H=\varnothing$, and $U\cup H=B$. I’ll show that $U$ and $H$ are separated sets, contradicting the connectedness of $B$.

Suppose that $x\in H$; then there is an open $V$ in $B$ such that $x\in V$ and $V\cap G=\varnothing$. Moreover, $x\in(B\setminus A)\setminus C$, and $C$ is a component of $B\setminus A$, so there is an open $W$ in $B$ such that $x\in W$ and $W\cap C=\varnothing$. Thus, $V\cap W$ is an open nbhd of $x$ disjoint from $U$. Since $x\in H$ was arbitrary, $H\cap\operatorname{cl}U=\varnothing$.

Now suppose that $x\in U$; then either $x\in G$, or $x\in C$. If $x\in G$, there is an open $V$ in $B$ such that $x\in V$ and $V\cap H=\varnothing$, so $x\notin\operatorname{cl}H$. If $x\in C$, there is an open $V$ in $B$ such that $x\in V$ and $V\cap H=V\cap\big(B\setminus(G\cup C)\big)\subseteq B\setminus\big(A\cup C\big)=V\cap\big((B\setminus A)\setminus C\big)=\varnothing$, since $C$ is open in $B\setminus A$, so again $x\notin\operatorname{cl}H$. Thus, $U\cap\operatorname{cl}H=\varnothing$, and $H$ and $U$ are separated sets. $\dashv$

Corollary: Suppose that $\{A_k:k=1,\dots,n\}$ is a finite family of connected sets such that $A_i\cup A_k$ is not connected whenever $1\le i<k\le n$; then $\bigcup_{k=1}^nA_k$ is not connected.

Proof: This is trivial for $n\le 2$, so suppose that $n\ge 3$, and let $S=\bigcup_{k=1}^nA_k$. Since $A_i\cup A_k$ is not connected whenever $1\le i<k\le n$, the sets $A_k$ are pairwise disjoint, and therefore $S\setminus A_1=\bigcup_{k=2}^nA_k$, whose connected components are the sets $A_k$ for $k=1,\dots,n-1$. It follows from (a) that $S\setminus A_1$ is connected, which is absurd: its components are the $A_k$ for $k=2,\dots,n$. $\dashv$

I will also need the following consequence of (a) and axiom (iii):

Proposition: If $A,B$, and $C$ are connected, and $A\cup B$ and $A\cup C$ are not connected, then $A\cup B\cup C$ is not connected. (Note that axiom (iii) is trivial to prove in the topological setting.)

Proof: Let $S=A\cup B\cup C$ and $D=B\cup C$, and suppose that $S$ is connected. I show first that $D$ is connected. Since $A\cup B$ and $A\cup C$ are not connected, $A\cap D=\varnothing$. Suppose that $D$ is not connected. Then $B\cap C=\varnothing$, and $B$ is a component of $D=S\setminus A$. By (a), therefore, $S\setminus B=A\cup C$ is connected, contrary to hypothesis, and it follows that $D$ must be connected.

Next, $A\cup D$ is connected, so by (iii) there is an $x\in A\cup D$ such that $A\cup\{x\}$ and $D\cup\{x\}$ are connected. If $x\in D$, then without loss of generality $x\in B$, and $A\cup B=(A\cup\{x\})\cup B$, where $(A\cup\{x\})\cap B\ne\varnothing$, so $A\cup B$ is connected, contrary to hypothesis. Thus, $x\in A\setminus D$, and a similar argument shows that $B\cup\{x\}$ and $C\cup\{x\}$ are not connected. Let $B\,'=B\cup\{x\}$. Then $B\,'\cup C=D$ is connected, so by (iii) there is a $y\in D$ such that $B\,'\cup\{y\}$ and $C\cup\{y\}$ are connected. $B\,'$ is not connected, so $y\in C\setminus B\,'$. Note that $\{x,y\}$ is not connected: if it were, then $C\cup\{x\}=C\cup\{x,y\}$ would be, since $C\cap\{x,y\}=\{y\}\ne\varnothing$.

Now $B\cup\{x,y\}=B\,'\cup\{y\}$ is connected, and $\{x,y\}$ is not, so by (iii) there is a $z\in B$ such that $\{x,y,z\}$ is connected. Then $\{x,y\}\cup\{x,z\}$ is connected, so by (iii) there is a $p\in\{x,y,z\}$ such that $\{x,y,p\}$ and $\{x,z,p\}$ are connected. $\{x,y\}$ is not connected, so $p=z$, $\{x,z\}$ is connected, and therefore $B\cup\{x,z\}$ is connected. But $B\cup\{x,z\}=B\,'$, which is not connected. This final contradiction shows that $S$ is not connected. $\dashv$

(b) If $\{A_k:k=1,\dots,m\}$ and $\{B_k:k=1,\dots,n\}$ are finite families of connected sets, and $\bigcup_{k=1}^mA_k\cup\bigcup_{k=1}^nB_k$ is connected, then there are $i\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$ such that $A_i\cup B_j$ is connected.

Proof: The result is certainly true if $m=n=1$. If it is not true for all $m,n\in\Bbb Z^+$, let $\{A_k:k=1,\dots,m\}$ and $\{B_k:k=1,\dots,n\}$ be a counterexample with $m+n$ as small as possible. At least one of $m$ and $n$ is greater than $1$; without loss of generality assume that $n>1$. Let $S=\bigcup_{k=1}^mA_k\cup\bigcup_{k=1}^nB_k$.

Suppose that $B_i\cup B_k$ is connected for some distinct $i,k\in\{1,\dots,n\}$; without loss of generality $\{i,k\}=\{n-1,n\}$. Then $\{A_k:k=1,\dots,m\}\cup\{B_k:k=1,\dots,n-2\}\cup\{B_{n-1}\cup B_n\}$ is a family of $m+n-1$ connected sets whose union is connected, and $A_i\cup B_k$ is not connected for any $i\in\{1,\dots,m\}$ and $k\in\{1,\dots,n-2\}$, so there is a $k\in\{1,\dots,m\}$ such that $T=A_k\cup B_{n-1}\cup B_n$ is connected, contradicting the proposition. Thus, $B_i\cup B_k$ is not connected whenever $1\le i<k\le n$. Similarly, $A_i\cup A_k$ is not connected whenever $1\le i<k\le m$, and the result follows from the corollary. $\dashv$

(c) If $\{A_k:k=0,\dots,n\}$ is a finite family of connected sets such that $\bigcup_{k=0}^nA_k$ is connected, then the sets $A_1,\dots,A_n$ can be re-indexed so that $A_0\cup\bigcup_{k=1}^mA_k$ is connected for $m=1,\dots,n$.

Proof: Applying (b) to the families $\{A_0\}$ and $\{A_1,\dots,A_n\}$, we see that there is some $k\in\{1,\dots,n\}$ such that $A_0\cup A_k$ is connected; re-index $\{A_1,\dots,A_n\}$ by interchanging the indices of $A_k$ and $A_1$, so that $A_0\cup A_1$ is connected. Now apply (b) to the families $\{A_0\cup A_1\}$ and $\{A_2,\dots,A_n\}$ to find $k\in\{2,\dots,n\}$ such that $A_0\cup A_1\cup A_k$ is connected, and re-index $\{A_2,\dots,A_n\}$ by interchanging the indices $2$ and $k$. In general, if $0\le m<n$, and $\bigcup_{k=0}^mA_k$ is connected, (b) ensures that there is an index $i\in\{m+1,\dots,n\}$ such that $A_i\cup\bigcup_{k=0}^mA_k$ is connected, and we re-index $\{A_{m+1},\dots,A_n\}$ by interchanging the indices $m+1$ and $i$. $\dashv$

Added: It’s easier to prove (b) in the purely topological setting; proving it using just (a) and the axioms in the cited paper made it a bit harder.

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In the last paragraph, it is asked whether $A\cup C$ is connected when $A\subseteq B$ are connected sets and $C$ is a component of $B\setminus A$. This is generally false; consider the planar space $B=\{c\}\cup\bigcup_{n=1}^{\infty}A_n$, where $c=\langle 1,0\rangle$, and $A_n=\{\langle x,\frac{x}{n}\rangle:0\leq x\leq 1\}$, $n=1,2,\dots$ Letting $A=\{\langle 0,0\rangle\}$ and $C=\{c\}$, we see that $C$ is a component of $B\setminus A$; however $A\cup C$ is a two-point discrete space. What IS generally true is that $A\cup C$ is connected, provided $C$ is clopen in $B\setminus A$.

What is also true is that if $A\subseteq B$ are continua (=connected+compact+Hausdorff) and $C$ is a component of $B\setminus A$, then $A\cup C$ is a continuum. To see this, first note that $C$ is closed in $B\setminus A$; hence any limit point of $C$ in $B$ must lie in $A$. With $\overline{C}$ denoting the closure of $C$ in $B$, we than have $A\cup C=A\cup\overline{C}$ is closed in $B$, and so compact Hausdorff. If $C$ were itself a continuum, then "boundary bumping" would ensure the existence of a continuum $K\subseteq B\setminus A$ that contains $C$ properly. This would kill $C$'s chances of being maximally connected in $B\setminus A$; hence we conclude that $\overline{C}$ intersects $A$, and therefore that $A\cup C$ is connected.

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