Dgdrxrt

Suppose $A_1,dots,A_k$ are connected open subsets of $[0,1]$ such that $[0,1]=bigcup_i=1^k A_i$ and $A_i notsubseteq A_j$ for each $ine j$.

By characterization of connected subsets of $mathbbR$ 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 $ileq k$ (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 $ine j$ such that $b_i=b_j$ otherwise we would have $A_isubseteq A_j$ or vicecersa.

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

I want to show that $$b_i+1<c_i$$ for each $i=1,dots,k-1$

For absurd assume that $b_i+1ge c_i$ and take $xin [c_i,b_i+1]$. Since $x ge c_i$ it follows that $xnotin A_j$ 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! :)