# Union of connected subsets is connected if intersection is nonempty

Let $\mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $\bigcap\mathscr{F}\ne\emptyset$. Prove that $\bigcup\mathscr{F}$ is connected.

If $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $\bigcap\mathscr{F}$. Then either $x\in A$ or $x\in B$. I don't know where to go from here.

HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $x\in\bigcap\mathscr{F}$, and without loss of generality assume that $x\in A$. $B\ne\varnothing$, so pick any $y\in B$. Then there is some $F\in\mathscr{F}$ such that $y\in F$, and of course $x\in F$. Thus, $x\in A\cap F$, and $y\in B\cap F$, so $A\cap F\ne\varnothing\ne B\cap F$. Why is this a contradiction?

• @TaylorRendon: No, the problem is that $A$ and $B$ split the connected set $F$. Oct 15, 2020 at 15:12
• @AriRoyceHidayat: Definitely not: $y$ can be any point of $\bigcup\mathscr{F}$. Jun 25 at 18:53
• @AriRoyceHidayat: $x\in\bigcap\mathscr{F}$, so $x\in F$ for every $F\in\mathscr{F}$. Jun 25 at 19:34
• @AriRoyceHidayat: You’re welcome. Jun 25 at 21:24
• @AriRoyceHidayat: If $A,B$, and $C$ are connected, $A\cap B\ne\varnothing$, and $B\cap C\ne\varnothing$, then $A\cup B\cup C$ is connected; it doesn’t matter whether $A\cap C$ is empty or not. From the result in the question we know that if $A\cap B\ne\varnothing$, then $A\cup B$ is connected. Let $D=A\cup B$; then $D\cap C\supseteq B\cap C\ne\varnothing$, and $D$ and $C$ are connected, so the same result tells us that $D\cup C$ is connected, i.e., that $A\cup B\cup C$ is connected. Aug 7 at 7:23

Use that $X$ is connected if and only if the only continuous functions $f:X\to\{0,1\}$ are constant, where $\{0,1\}$ is endowed with the discrete topology.

Now, you know each $F$ in $\mathscr F$ is connected. Consider $f:\bigcup \mathscr F\to\{0,1\}$, $f$ continuous.

Take $\alpha \in\bigcap\mathscr F$. Look at $f(\alpha)$, and at $f\mid_{F}:\bigcup \mathscr F\to\{0,1\}$ for any $F\in\mathscr F$.

Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:

THM Let $(X,\mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:X\to\{0,1\}$ is continuous, it is constant. The space $\{0,1\}$ is endowed with the discrete metric (topology), that is, the open sets are $\varnothing,\{0\},\{1\},\{0,1\}$.

P First, suppose $X$ is disconnected, say by $A,B$, so $A\cup B=X$ and $A\cap B=\varnothing$, $A,B$ open. Define $f:X\to\{0,1\}$ by $$f(x)=\begin{cases}1& \; ; x\in A\\0&\; ; x\in B\end{cases}$$

Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in $\{0,1\}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:X\to\{0,1\}$ is continuous but not constant. Set $A=\{x:f(x)=1\}=f^{-1}(\{1\})$ and $B=\{x:f(x)=0\}=f^{-1}(\{0\})$. By hypothesis, $A,B\neq \varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $A\cup B=X$ and $A\cap B=\varnothing$. Thus $X$ is disconnected. $\blacktriangle$

I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a \in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $\emptyset \neq C \subsetneq X$.

Take $a \in \bigcap \mathscr{F}$. Then, take a clopen set $C \subset \bigcup \mathscr{F}$ containing $a$. In the relative topology, for any $X \in \mathscr{F}$, $C \cap X$ is a clopen set. Since $X$ is connected, $C \cap X = X$. That is, $X \subset C$ for every $X \in \mathscr{F}$. Therefore, $\bigcup \mathscr{F} \subset C$.

Therefore, $\bigcup \mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.

The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X \in \mathscr{F}$. From connectedness, we have that $X \subset C'$. In particular, $a \in C'$.

The quickest way is using functions from $\bigcup\mathcal{F}$ to $\{0,1\}$, However you can do it directly.

Let $A, B$ be a partition of $\bigcup\mathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $x\in A$ and hence some $F\in \mathcal{F}$ so that $x\in F$. As $F$ is connected it follows that $F\subseteq A$, hence $\bigcap \mathcal{F} \subseteq F \subseteq A$. As this intersection is non-empty every $G\in \mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $G\subseteq A$. Hence every member of $\mathcal{F}$ is a subset of $A$ and so $B$ is empty.

Denote the whole topology space as $$(\bigcup \mathcal{F},\mathcal{T})$$.

Suppose $$\bigcup \mathcal{F}$$ is not connected, then there exists disjoint two non-empty open sets $$A,B$$ that decompose $$\bigcup \mathcal{F}$$, that is , $$A,B \neq \emptyset$$, $$A \bigcap B = \emptyset$$, $$A \bigcup B = \bigcup \mathcal{F}$$ and $$A,B\in\mathcal{T}$$.

Since $$\bigcap \mathcal{F} \neq \emptyset$$, so $$\exists x \in \bigcap \mathcal{F}$$, W.L.O.G we assume $$x \in A$$.

Meanwhile, since $$B \neq \emptyset$$, we know $$B \bigcap \mathcal{F}_j \neq \emptyset$$ for some $$\mathcal{F}_j$$, and $$\exists y \in B \bigcap \mathcal{F}_j$$.

Then we claim that $$\mathcal{F}_j$$ can be decomposed into two non-empty disjoint open sets $$A \bigcap \mathcal{F}_j$$ and $$B \bigcap \mathcal{F}_j$$. This is because, $$x \in A \bigcap \mathcal{F}_j \Rightarrow (A \bigcap \mathcal{F}_j) \neq \emptyset$$; $$y \in B \bigcap \mathcal{F}_j \Rightarrow (B \bigcap \mathcal{F}_j) \neq \emptyset$$;
$$A \bigcap B = \emptyset \Rightarrow (A \bigcap \mathcal{F}_j) \bigcap (B \bigcap \mathcal{F}_j) = \emptyset$$;
$$A \bigcup B = \bigcup \mathcal{F} \Rightarrow (A \bigcap \mathcal{F}_j) \bigcup (B \bigcap \mathcal{F}_j) = (A \bigcup B) \bigcap \mathcal{F}_j = \mathcal{F}_j$$;
$$A,B \in \mathcal{T} \Rightarrow (A \bigcap \mathcal{F}_j), (B \bigcap \mathcal{F}_j) \in \mathcal{T}_{F_j}$$, where $$(\mathcal{F}_j,\mathcal{T}_{F_j})$$ is the induced subspace.

The above claim contradicts with the assumption that each $$\mathcal{F}$$ is connected, hence we know $$\bigcup \mathcal{F}$$ is connected.

We begin with a Lemma:

If $$X$$ is a topological space with separation $$A \cup B$$ and $$Y \subset X$$ is connected, then $$Y \subset A$$ or $$Y \subset B$$. The proof goes by noticing that $$(Y \cap A) \cup (Y \cap B)$$ is a disconnection of $$Y$$. Which cannot happen as $$Y$$ is connected.

Now for main proof: Let us suppose $$\bigcup \mathcal{F} = A \cup B$$ where $$A,B$$ are disjoint open nonempty subsets of our space $$M$$. Then there exists some $$x \in \bigcap \mathcal{F}$$. WLOG say $$x \in A$$. But $$B$$ is nonempty thus there exists some $$b \in B$$ forcing $$b \in \mathcal{F}_\beta$$ for some $$\beta$$ in our indexing set. But as $$x \in \bigcap \mathcal{F}$$ we have $$x \in \mathcal{F}_\beta$$ but by our lemma $$\mathcal{F}_\beta \subset A$$ or $$\mathcal{F}_\beta \subset B$$ thus the union is connected.

Suppose that $$\bigcup\mathscr{F}$$ is not connected, then it can be partitioned into two disjoint, non-empty, open subsets $$A$$ and $$B$$.

As $$\forall F \in \mathscr{F}$$ are connected, there's no need to investigate whether $$A$$ and $$B$$ would disconnect the union in any of them. So, without loss of generality, let's define $$X = \{x : x \in \bigcap \mathscr{F} \cap A\}$$ and $$Y = \{y : y \in \bigcap \mathscr{F} \cap B\}$$. Since $$A$$ and $$B$$ are disjoint, it implies $$X \cap Y = \emptyset$$.

It is a contradiction because $$\bigcap \mathscr{F} \ne \emptyset$$. Thus $$\bigcup\mathscr{F}$$ must be connected.

• Comment to Brian M. Scott's answer: so $A′ = A \cap F \ne \emptyset$, $B′ = B \cap F \ne \emptyset$, $A′ \cap B′ = \emptyset$ and $A′ \cup B′ = F$. It is a contradiction because F is connected. Aug 7 at 17:07