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

I am having difficulty understanding the proof in my book, which goes:

Suppose that $A,B$ are open sets forming a disconnection of $\mathbb{I}$. Thus $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$ are non-empty bounded disjoint sets whose union is $\mathbb{I}$. Since $A$ and $B$ are open, the sets $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$ cannot consist of only one point. (Why?) For the sake of definiteness, we suppose that there exist points $a\in A$, $b\in B$ such that $0<a<b<1$. Applying the supremum property, we let $c=\sup\{x\in A:x<b\}$ so that $0<c<1$; hence $c\in A\cup B$. If $c\in A$ then $c\ne b$ and since $A$ is open there is a point $a_1\in A$, $c<a_1$, such that the interval $[c,a_1]$ is contained in $\{x\in A: x<b\}$ contrary to the definition of $c$.

Why is $0<c<1$? They define $c=\sup\{x\in A:x<b\}$ so it is the supremum of $A$ which is less than $b$ so doesn't that mean $c$ must be less than $b$ ? Also, why is $c\in A\cup B$ ? Lastly, since $c$ is the supremum of $A$ how is there a point in $A$ such that $c<a_1$ and why is it contrary that it is contained in $\{x\in A: x<b\}$ ?

  • $\begingroup$ Isn't interval defined as connected set (in standard real topology)? $\endgroup$ – jdh8 Oct 1 '15 at 10:24
  • $\begingroup$ No. A non-empty set $J$ of reals is called an interval iff for every two elements of $J$ all elements between them belong to $J$, too. $\endgroup$ – Michael Hoppe Dec 14 '17 at 15:41

$A$ meets $\mathbb{I}$ non-trivially so $c$ is bigger than $0$ necessarily. Since $b<1$ and $b$ is an upper bound for $\{ x \in A : x < b \} $, $c \leq b < 1$. So, $0<c<1$.

It follows that $c \in (0,1) \subseteq [0,1] = A \cup B$ by supposition.

Now we'll assume that $c \in A$.

Necessarily $c < b$.

If $c=b$ then $c \in A \cap B$ however $A \cap B = \varnothing$ and also $c$ cannot exceed $b$.

We can pick some $\varepsilon_1$ neighborhood around $c$ that is contained in $A$ since it is an open set. Pick some $c< a_1< c+\varepsilon_1$ in that neighborhood. $A$ and $B$ are disjoint so this neighborhood will not contain $b$ so this $a_1 < c+\varepsilon_1 < b$ so $a_1 \in \{ x \in A : x < b \}$ but then $a_1 \leq c$ since it is the supremum but also we have $a_1 > c$, a contradiction.

  • $\begingroup$ Thanks a lot! One question, in the last paragraph we can pick a neighborhood with an open ball which is great than $c$ but still contained in $A$ because of the fact that $A$ is open, correct? Which is why a $a_1 >c$? $\endgroup$ – Q.matin Mar 24 '13 at 20:31
  • $\begingroup$ Yes correct, $A$ is open so we can put a neighborhood around $c$ that is contained in $A$. The neighborhood looks like: $\{ x \in \mathbb{R} : x \in (c-\varepsilon_1,c+\varepsilon_1) \}$ so can just pick some $a_1$ that is larger than $c$ but still contained in this neighborhood. I also edited my answer as I didn't need to pick another epsilon heh. $\endgroup$ – Frudrururu Mar 24 '13 at 23:13
  • $\begingroup$ Understand, thanks a lot ! $\endgroup$ – Q.matin Mar 25 '13 at 3:17

For the first question, consider $S = \{ x \in A : x < b \}$. Now $b$ is an upper bound. As given, $0 < b < 1$. As $c$ is the least upper bound of $S$, then $0 \le c < b$. Since $b < 1$, then $0 \le c < 1$. Now, why can't $c$ equal $0$? If that were so, then $S = \{0\}$, but then that means $A$ has $0$ isolated, but that can't be so, since $A$ is open, and $[0, b)$ is open (in $[0, 1]$), and then we'd have that $A \cap [0, b) = S = \{0\}$, and $\{0\}$ is closed (being the complement of $(0, 1]$, which is open in $[0, 1]$). But the intersection of two open sets is always open.

Why $c \in A \cup B$? Well, since $0 < c < 1$ then $c \in (0, 1)$. But $A \cup B = [0, 1]$, and $(0, 1) \subset [0, 1]$. So $c \in A \cup B$.

Why such an $a_1$? As $A$ is open, then there is an $\epsilon$-ball centered at $c$ contained in $A$. So it'll contain some points greater than $c$. And if $[c, a_1] \subset S$, then we'd have $c < a_1$, yet $a_1 \in S$, but that contradicts that $c$ was an upper bound of $S$.

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    $\begingroup$ I'm not sure your argument is valid. $[0,a)$ is open in the subspace topology on $[0,1]$. Then an example of an open neighbourhood of $0$ would be some $[0,\varepsilon)$ and similarly $(1-\varepsilon,1]$ for $1$. $\endgroup$ – Ian Coley Mar 24 '13 at 7:19
  • $\begingroup$ Thanks Mike! What do you mean by, "This is because any ϵ -ball centered at 0 or 1 would necessarily contain points not in A or B , no matter how small the ϵ"? Also, why is $A \cup B = [0, 1]$ I thought you said in the first sentence that they cannot include $0$ or $1$? $\endgroup$ – Q.matin Mar 24 '13 at 7:24
  • $\begingroup$ @Frank McGovern: How's this new argument? $\endgroup$ – The_Sympathizer Mar 24 '13 at 7:39
  • $\begingroup$ @Q matin: Changed that bit. $\endgroup$ – The_Sympathizer Mar 24 '13 at 7:46

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