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In the book of The elements of Real Analysis by R. Bartle, at page 78, it is given that

The closed unit interval $I= [0,1]$ is a connected subset of $\mathbb{R}$.

Proof:

We proceed by contradiction and suppose that $A,B$ are open sets forming a disconnection of $I$.Thus, $A\cap I$ and $B \cap I$ are non-empty bounded disjoint sets whose union is $I$.WLOG, assume $1\in B$, then let $c = sup A\cap I$ so that $c\in A\cup B$. If $c\in A$, then $0<c<1$; since $A$ is open and contains $c$, there are points in the intersection, hence contradiction. If $c\in B$, then $0<c \leq 1$; there is a point $c_1 < c$ s.t $[c_1, 1] \subseteq B\cap I$, which also contradicts the definition of c being supremum of $A\cap I$.

First of all, if $c \in sup(A\cap I)$, how can $c\in B$ ?

Secondly, if $c \in B$, why do we get a contradiction by showing $[c_1,1] \subseteq B$ ?

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marked as duplicate by Andrés E. Caicedo, Jack, Community Dec 14 '17 at 15:21

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  • $\begingroup$ The fact is that it can't, because the sets $A$ and $B$ can't even exist. But under the assumption, we have $c\in A\cup B$, so $c\in A$ or $c\in B$. $\endgroup$ – ajotatxe Dec 14 '17 at 15:13
  • $\begingroup$ @ajotatxe However, the definition of $c = sup A \cap I$, and $(A\cap I) \cap (B \cap I) = \emptyset$. Hence shouldn't we have $c\in A$, hence $c\not \in B$ ? $\endgroup$ – onurcanbektas Dec 14 '17 at 15:16
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    $\begingroup$ The supremum of a set may or may not belong to it. So it could be the fact that $c\notin A \cap I$ and so it has to belong to $B$. $\endgroup$ – Alejandro Nasif Salum Dec 14 '17 at 15:16
  • $\begingroup$ What is the reason for the close vote ? $\endgroup$ – onurcanbektas Dec 14 '17 at 15:16
  • $\begingroup$ @AlejandroNasifSalum You are right, I have forgotten that. Thanks for pointing out. $\endgroup$ – onurcanbektas Dec 14 '17 at 15:17
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You say "$c \in sup(A\cap I)$", but really, it's $c = \sup(A\cap I)$. The supremum of a set of numbers is not a set of numbers, it's a number. As for how we can have $c\in B$, $A$ and $B$ together make up $I$, so $c$ has to be in one of them. However, a set needs not contain its $\sup$, so if $c\notin A\cap I$, then we must have $c\in B$.

The contradiction in the $c\in B$ part comes from the combination of $c_1<c\leq 1$, and $[c_1,1]\subseteq B$. It means that $c_1$ is a smaller upper bound of $A\cap I$ than $c$ is, which contradicts the definition of $\sup$.

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  • $\begingroup$ Actually I have written $c=sup A \cap I$ intentionally because this is what the author write in the book, so I directly quoted him. $\endgroup$ – onurcanbektas Dec 14 '17 at 15:20
  • $\begingroup$ In the quote, yes. Below the quote from the book, however, you've written "First of all, if $c\in sup(I\cap A)$", which is what I'm referring to. $\endgroup$ – Arthur Dec 14 '17 at 15:21
  • $\begingroup$ Oh, I see. I did not notice. $\endgroup$ – onurcanbektas Dec 14 '17 at 15:24
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    $\begingroup$ @onurcanbektas I guess it was just a typo then. That's fine (I had an entirely analoguous mistake in my master's thesis, mixing up $\in$ and $\subseteq$, so it happens to all of us). It's just that with $\sup$ it's an actual misconception among some students who aren't yet entirely comfortable with the concept of suprema, so I felt it was worth it to make sure, just in case. $\endgroup$ – Arthur Dec 14 '17 at 15:26
  • $\begingroup$ I see, thanks for that. $\endgroup$ – onurcanbektas Dec 14 '17 at 16:14

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