A quick question about Theorem 24.1 of Munkres.

Question

Can someone please help me understand why the underlined statement must be true. Why must there be a $d$ in $B_0$ less than $c$? Why can't $c$ be the smallest element of $B_0$?

Answer 1

Note that $B_0$ is open in the subspace topology on $[a,b]$, which is the same as the order topology on $[a,b]$ because $[a,b]$ is a convex subset of $L$ (see Theorem 16.4 in the book). Therefore the subspace topology on $[a,b]$ is generated by the collection $\mathcal{B}$ of all sets of the following types:

$(1)$ All open intervals $(x,y)$ with $x<y$ and $x,y\in[a,b]$.

$(2)$ All intervals of the form $[a,x)$ with $x\in(a,b]$.

$(3)$ All intervals of the form $(x,b]$ with $x\in[a,b)$

Now, $c\in B_0$ and $B_0$ open in $[a,b]$ implies that there exists $B\in\mathcal B$ such that $c\in B\subset B_0$. There are three cases:

Case $B$ is of type $(1)$. Then $(x,c]\subset B\subset B_0$ for some $x\in[a,c)$.

Case $B$ is of type $(2)$. Then $[a,x)\subset B_0$ for some $x\in(a,b]$, which is a contradiction since $a\notin B_0$. Therefore this case is impossible.

Case $B$ is of type $(3)$. Then again $(x,c]\subset B\subset B_0$ for some $x\in[a,c)$.

We conclude that there exists $x\in[a,c)$ such that $(x,c]\subset B_0$, as desired.

Answer 2

Because $B_0$ is open in $[a,b]$ and $c\in B_0$, by definition of open set, there exists a basic of $[a,b]$ containing c that contained in $B_0$, the basis of $[a,b]$ is of the form $(x, y) \cap [a, b]$ where $x,y \in L$. Hence there e exists $d$ suct that $(d, c] \in B_0$.

Answer 3

Since $c\in B_0=B\cap[a,b]$ and $B$ is open in $Y$, then $B$ is the union of (possibly infinitely many) open intervals and/or rays. Since $c\in B\cap[a,b]$, then $c$ is in some interval of the form $(x,y)\cap[a,b]$ that is contained in $B\cap [a,b]$. Hence there is some $d$ so that $(d,c]\subseteq B_0$ which is non-empty because, first, $c\in(d,c]$ and second, $L$ is a linear continuum.