Htykuut

Let $mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $bigcapmathscr{F}neemptyset$. Prove that $bigcupmathscr{F}$ is connected.

If $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $bigcapmathscr{F}$. Then either $xin A$ or $xin B$. 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 $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?

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

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

Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.

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 $f:Xto{0,1}$ 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 $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$

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 $f:Xto{0,1}$ 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, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. 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 $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.

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

In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.