Exercise 17.5

Proof. Since $(a,b)\subseteq [a,b]$ closed, and by the definition of closure,

$\overline{(a,b)}=[a,b]⟺a,b\in \overline{(a,b)}⟺$ any basis elements $A\ni a,B\ni b$ intersect $(a,b)$ by Theorem $17.5(b)$. We claim that this is equivalent to the fact that there is no immediate successor $\alpha$ of $a$ and no immediate predecessor $\beta$ of $b$. If either are the case, say for $a$, then choosing $A$ with upper bound $\alpha$ would not intersect $(a,b)$, and so equality doesn’t hold since $a\notin \overline{(a,b)}$; in the other direction, if neither are the case, we see that, say for $a$, the upper bound of $A$, $\alpha$ would be such that $(a,\alpha )$ is non-empty, and so $A\cap (a,b)\ne \varnothing$, satisfying the condition for Theorem $17.5(b)$. The same argument applies when considering $b$ and $\beta$, and so our claim holds. □

Proof. First, the closed interval $[a,b]$ is closed (hence why it is called such!) because clearly, its complement is

and we know that open rays are always open so their union is as well. Clearly also $[a,b]$ contains $(a,b)$. Hence $[a,b]$ is a closed set containing $(a,b)$. Since $\overline{(a,b)}$ is defined as the intersection of closed sets that contain $(a,b)$ clearly we have that $\overline{(a,b)}\subset [a,b]$ as desired. □

Proof. $\text{(⇒)}$ We show the contrapositive of this. So suppose that $a$ does have an immediate successor $c$. Then the open ray $(-\infty ,c)$ is an open set that contains $a$ but does not intersect $(a,b)$. This is easy to see, because if they did intersect, there would be an $x\in (a,b)$ where also $x\in (-\infty ,c)$. From these, it follows that $a<x<c$, which contradicts the fact that $c$ is the immediate successor of $a$. Hence by definition, $a$ is not a limit point of $(a,b)$.

$\text{(⇐)}$ Suppose that $a$ is not a limit point of $(a,b)$. Then there is an open set $U$ containing $a$ that does not intersect $(a,b)$. From this, it follows that there is a basis element $B$ containing $a$ such that $B\subset U$, and thus $B$ also cannot intersect $(a,b)$ (as, if it did, then so would $U$). Suppose that $B$ is the open interval $(c,d)$ so that $c<a<d$. It also must be that $d<b$ for otherwise, for any element of $x$ of $(a,b)$, we would have $c<a<x<b\le d$ so that $x\in (c,d)=B$ and $B$ and $(a,b)$ would not be disjoint. We claim that $d$ is the immediate successor of $a$. If this is not the case then there would be an $x$ such that $c<a<x<d$ and hence $x\in (c,d)=B$. Also $a<x<d<b$ so that also $x\in (a,b)$. Therefore $B$ and $(a,b)$ would not be disjoint. Similar arguments can be made if $B$ are other types of basis element in the order topology. (Actually, $B$ cannot be of the form $(e,f]$ for the largest element $f$ of $X$ since then any element of $(a,b)$ would also be in $B$ and they would not be disjoint.) □