Understanding the proof of "connected set is interval."

Question

There is some questions about connected set.

The first question arises in logical translation. I translated the property of interval into logical proposition that $\forall a,b \in E, \forall c : [a<c<b \rightarrow c \in E]$.

Is it right? I think it is something wrong. If there is $d$ which satisfies $a<b<d$ or is not able to order $\{a,c,d\}$, then the assumption of logical proposition that $a<d<b$ is false so the statement is vacously true. What is good translation of the property of interval?

The second arises in proving "Connected set is interval". The following is definition of disconnected set, and connected set is defined by not being disconnected.

\begin{align} E \subseteq \mathbb{R} \textrm{ is disconnected }\quad \Leftrightarrow \quad & \exists A,B \textrm{ which are open sets such that } \\ &(i) A \cap E \not = \varnothing, B \cap E \not = \varnothing \\ &(ii) (A \cap E) \cap (B \cap E) = \varnothing \\ &(iii) (A \cap E) \cup (B \cap E) = E \end{align}

The following is the proof of "if $E \subseteq \mathbb{R}$ is interval, then $E$ is connected set."

Suppose not. Since $A \cap E \not = \varnothing, B \cap E \not = \varnothing$, there exists $a \in A \cap E$, $b \in B \cap E $. By trichotomy property , $a=b, a<b, a>b$. We defined $A \cap E, B \cap E$ disjoint, so $a \not = b$. WLOG suppose $a<b$. Let $S = A \cap [a,b]$. For $S \subseteq [a,b]$, S is also bounded above, and there exists supremum S such that $c:=\sup S, a<c \leq b$.

Claim : $c \not = b$.

Suppose not. Then $c$ should be a interior point of $B$.($\because B$ is open set). Then there exists something suitable open ball with its center $c$ and radius $\epsilon >0$ : $B_\epsilon (c) \subseteq B$. Then $c- \epsilon/2$ is also upper bound of $S$, so contradiction.

I have a question in the last paragraph of this proof. How can we verify $c- \epsilon /2$ is also upper bound? What I think is to make it sense, the premise should be proved that $\forall a \in A \cup E, \forall b \in B \cup E \rightarrow a < b$. However my text doesn't prove it and just use this premise implicitly. Explain me about this.

Answer 1

From the definitions of $S$ and $c$ we have $A \cap [a,b] \subset [a,c]$ but $(c- \epsilon /2 ,c] \subset B$ and $A \cap B = \phi$ which gives $S=A \cap [a, b] \subset [a,c- \epsilon /2]$. This last expression implies that $c- \epsilon /2$ is an upper bound for $S$.

Answer 2

The property you mention is a nice way to deal with all possible intervals from the real line, for it captures exactly the intervals, no matter if they are unbounded, half open etc. You could read the property as $E$ contains each point which lies between any two of its points.

As for some $d$ such that $a<b<d$ it indeed falsifies the premise $a<d<b$ and so for the implication $(a<d<b) \rightarrow (d \in E)$ to be true, not necessarily $d \in E$.

As for the proof, I think what is needed to be proved is that $c \not \in A$ as well as $c \not \in B$, right? If this is the case, then $E$ is not an interval because $c$ lies between two of its points ($a$ and $b$) but $c\not \in E$.

To see that $c \not \in A$, openness is needed. If $c \in A$, then there would be an open ball $B_\epsilon(c) \subseteq A$. But then for all $x \in B_\epsilon(c)$ such that $c < x < c+\epsilon$, also $x \in [a,b]$ what means that $x \in A \cap [a,b]$. This contradicts the fact that $c$ is an upper bound to $A \cap [a,b]$.

To see that $c \not \in B$, suppose it does and let $B_\epsilon(c) \subseteq B$ be an open ball. Then consider any $x \in B_\epsilon(c)$ such that $c-\epsilon < x < c$. There is no point of $A \cap [a,b]$ greater or equal than $c+\epsilon$ for this would contradict the fact that $c$ is one of its upper bounds. Also $B_\epsilon(c)$ is disjunct from $A$ as it is a subset of $B$. Therefore all such $x$ must be an upper bound for $A \cap [a,b]$ which contradicts the fact that $c$ is its least upper bound.