Intuitively, how is the proof of IVT with connectedness equivalent to the proof with completeness?

Question

I just learned the proof of IVT in general topology and it seems very different than the one I learned in Analysis (actually, the Analysis IVT is a bit stronger, I guess, since it asserts that the point occurs in some interval). Does this mean that connectedness somehow encompasses the notion of completeness?

Answer 1

Let $<$ be a linear order on a set $S$. For $s,t∈S$ let $(s,t)=\{u∈S:s<u<t\}.$ Let $(←,s)=\{u∈S:u<s\}.$ Let $(s,→)=\{u∈S:u>s\}.$... Now let $T_<$ be the topology on $S$ generated by the base (basis) $\{(s,t):s,t∈S\}∪\{(←,s):s∈S\}∪\{(s,→):s∈S\}.$... Now suppose also that $<$ is a dense order: That is, if $s<t$ then $(s,t)≠∅.$

Then the relationship between connectedness of $T_<$ and order-completeness of $<$ is that $T_<$ is connected iff $<$ is order-complete.

Proof: (1).If $<$ is not order-complete then $T_<$ is a disconnected topology: Suppose $\emptyset \ne U\subset S$ where $U$ has an upper bound but no $lub.$ Let $V$ be the set of upper bounds for $U.$ For any $v\in V$ there exists $v'\in V$ with $v'<v$ . It should be obvious that $(v',\to)\subset V .$ So for any $v\in V$ there is an open set $(v',\to)$ such that $v\in (v',\to )\subset V.$ So $V$ is open.

Now let $V^*=\cup\{(\leftarrow,u):u\in U\}.$ Then $V^* $ is open. If $x\in V^*$ then $x<u$ for some $u\in U$ so $x\not \in V$, by def'n of $V$. And if $y\in V$ then $\neg (y<u)$ for all $u\in U$ so $y\not\in V^* .$

So $V\cap V^*=\emptyset.$

We have $V\ne \emptyset$ by hypothesis. And $U$ has no $\max$ because $U$ has no $lub,$ so if $u\in U$ there exists $u'\in U$ with $u\in (\leftarrow,u')\subset V^*.$ So $\emptyset \ne U\subset V^*.$

Finally, $V\cup V^*=S .$ Because if $x\in S\setminus V $ then by def'n of $ V$ there exists $u\in U$ with $x<u,$ so $x\in (\leftarrow,u)\subset V^*.$ So $V, V^*$ are disjoint, open, and non-empty, and their union is $S.$

Therefore $T_<$ is a disconnected topology.

By a similar method, if $\emptyset\ne U\subset S$ where $U$ has a lower bound but no $glb,$ then $T_<$ is disconnected. (Or we can apply the above argument verbatim to the order $<^*$, where $x<^*y\iff y<x,$ because $T_<=T_{<^*}$).

(2). If $<$ is order-complete AND if $<$ is order-dense then $T_<$ is a connected topology: We will use the following property $(^*)$ of $T_<$ when $<$ is order-dense:

(*). If $x<y$ then $x$ is in the closure of $(x,y).$

Suppose $S=A\cup B$ where $A,B$ are open and disjoint, and suppose that $A\ne \emptyset.$ Let $\alpha$ be some (any) member of $A.$ Suppose by contradiction that $ B\ne \emptyset$ and that $\alpha<\beta \in B.$ (The case $\alpha>\beta \in B$ can be handled similarly).

Let $W= A\cap (\leftarrow,\beta)$ and $\gamma=\sup W.$

We have $\gamma \in Cl(W) \subset Cl(A)=A.$

Now $\alpha\leq \gamma \leq \beta\in B$ and $\gamma \in A$ so $\gamma <\beta.$ And by def'n of $\gamma,$ the set $(\gamma,\beta)$ is a subset of $B.$ So by the property (*) we have $\gamma \in Cl(\,(\gamma,\beta)\,)\subset Cl (B)=B.$

So $\gamma \in A\cap B,$ a contradiction.