Connectedness of a linear continuum [Sect-24.1, Munkres]

Question

In Munkres' Topology p.153, we have a proof like this

Proof. $\ \ $ Recall that a subspace $Y$ of $L$ is said to be convex if for every pair of points $a,b$ of $Y$ with $a<b$, the entire interval $[a,b]$ of points of $L$ lies in $Y$. We prove that if $Y$ is a convex subspace of $L$, then $Y$ is connected.
$\quad$ So suppose that $Y$ is the union of the disjoint nonempty sets $A$ and $B$, each of which is open in $Y$. Choose $a\in A$ and $b\in B$: suppose for convenience that $a<b$. The interval $[a,b]$ of points of $L$ is contained in $Y$. Hence $[a,b]$ is the union of the disjoint sets $$A_0=A\cap[a,b]\quad{\rm and}\quad B_0=B\cap[a,b],$$ each of which is open in $[a,b]$ in the subspace topology, which is the same as the order topology. The sets $A_0$ and $B_0$ are nonempty because $a \in A_0$ and $b\in B_0$. Thus, $A_0$ and $B_0$ constitute a separation of $[a,b]$.
$\quad$ Let $c=\sup A_0$. We show that $c$ belongs neither to $A_0$ nor to $B_0$, which contradicts the fact that $[a,b]$ is the union of $A_0$ and $B_0$.
$\quad$ Case 1. Suppose that $c\in B_0$. Then $c\neq a$, so either $c=b$ or $a<c<b$. In either case, it follows from the fact that $B_0$ is open in $[a,b]$ that there is some interval of the form $(d,c]$ contained in $B_0$. If $c=b$, we have a contradiction at once, for $d$ is a smaller upper bound on $A_0$ than $c$. If $c<b$, we note that $(c,b]$ does not intersect $A_0$ (because $c$ is an upper bound on $A_0$). Then $$(d,b]=(d,c]\cup (c,b]$$

He mentions that $B_0$ is open, so there is some interval $(d, c]$ containing $c$, which is contained in $B_0$.

So we know if $c=b$, $(d, c]=(d, -\infty)$, which is open in $[a, b]$ for sure.

But if $a< c< b$, will $(d, c]$ still be open in this linear order?

My conclusion is $(d, c]$ is not open, because we assume in advance that

Definition. $\ \ $ A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

1. $L$ has the least upper bound property.

2. If $x<y$, there exists $z$ such that $x<z<y$.

Theorem 24.1. $\ \ $ If $L$ is a linear continuum in the order topology, then $L$ is connected, and so are intervals and rays in $L$.

L is a linear continuum, and for any $x$ and $y$, we can pick a $z$ lying between them. So if $(d, c]$ is open, we can write $(d, c]$ as $(d, e)$, but by the 2nd property of linear continuum, we can not have a number which is the minimal number in $[a, b]$, having the property "larger than $c$" (like what we do to prove $(a, x]$ is open is $\omega_1$). Am I correct? Thanks!

Answer 1

It's a fairly standard thing that comes up in topological arguments regarding "nice" types of spaces. Often there are certain types of useful not open sets, but where nevertheless every point $x \in U$ for $U$ open is contained in a set of this type contained in $U$. For example: strictly speaking an open set of $\mathbb{R}^n$ is one such that every $x$ has a small open ball $B$ around it such that $x \in B \subset U$. But if we halve the radius, we could also assume that the ball is closed, which can be helpful because of compactness, while still keeping the property $x \in B \subset U$.

In this example, the observation is that if $B_0 \subset [a,b]$ is open, then usually this means for $c \in B_0$ , if $c \in (a,b)$ we have some interval $c \in (d,f)$ about $c$. But this will also contain $(d,c]$, so every point has some half open interval around it. And this is nice because if we make an argument only using this, we avoid being nit-picky about the possibility $b=c$.

Answer 2

Because $ B_0 = A ∩ [a,b]$, where $A$ is an open set in $Y$, $B_0$ is an open set in $[a,b]$.

And, since $b\in B_0$, $B_0$ can be written in form $(t, b]$ for some $t$, $a<t<b$. Again since $B_0$ is an open set it contains a neighborhood (interval) of the form $(d, c]$, where $ t<d<c≤b$.