Suppose $A_1,\dots,A_k$ are connected open subsets of $[0,1]$ such that $[0,1]=\bigcup_{i=1}^k A_i$ and $A_i \not\subseteq A_j$ for each $i\ne j$.

By characterization of connected subsets of $\mathbb{R}$ I know that each $A_i$ is an interval.

I want to show:

There exist $0=a_0<a_1<\dots<a_k=1$ real numbers such that $[a_{i-1},a_i]\subseteq A_i$ for each $i\leq k$ (possibly permuting the $A_i$'s)

My argument

Permuting the $A_i$'s if necessary let's suppose

$$A_1=[0=b_1,c_1), A_2=(b_2,c_2),\dots, A_{k-1}=(b_{k-1},c_{k-1}), A_k=(b_k,c_k=1]$$

with the ordering $b_i<b_{i+1}$ for each $i=1,\dots,k-1$.

Note that it is not possible that there are indices $i\ne j$ such that $b_i=b_j$ otherwise we would have $A_i\subseteq A_j$ or vicecersa.

Note also that we have also $c_i<c_{i+1}$ for each $i=1,\dots,k-1$ otherwise we would have $A_{i+1}\subseteq A_i$.

I want to show that $$b_{i+1}<c_i$$ for each $i=1,\dots,k-1$

For absurd assume that $b_{i+1}\ge c_i$ and take $x\in [c_i,b_{i+1}]$. Since $x \ge c_i$ it follows that $x\notin A_j$ for each $j=1,\dots,i$. Since $x \leq b_{i+1}$ it follows that $x \notin A_j$ for each $j=i+1,\dots,k$. This is a contradiction since $[0,1]=\bigcup_{i=1}^k A_i$.

Now the proof ends by chosing $a_i \in (b_{i+1},c_i)$.

Is my proof correct?

P.S. Yesterday I posted a similar question Proof verification about a property of the topological space $[0,1]$ and it turned out that not only the proof was wrong, but also the statement. Even if now I have more carefully written the proof, my experience says that the errors are always lurking, so I will be happy if you control it. Thank you! :)


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