Union of connected subsets is connected if intersection is nonempty

Question

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.

Answer 1

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?

Answer 2

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 $F\in\mathscr F$.

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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. $\blacktriangle$

Answer 3

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.

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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, $a \in C'$.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.