Consider the basis $B = \{(a,\infty),a \in \mathbb{R}\}$ for a topology in $\mathbb{R}$. I've already proven that this is a basis, verifying that its union is the whole $\mathbb{R}$ and for every $x \in B_1 \cap B_2$, there is $B_3 \in B$ such that $x \in B_3 \subseteq B_1 \cap B_2$.
Now I have to describe the interior and closure operations in this topology.
Let $A=(a,\infty)$ (Why is it sufficient to analyze for a set $A$ in this form?)
Interior is the union of all open sets contained in $A$, and since $\tau = \{\bigcup_{(a,\infty) \in \cal{U}} | \cal{U} \in B\}$ is the topology, then $A^o=A=(a,\infty)$. (This looks fine to me, is it correct?)
Closure is the intersection of all closed sets that contain $A$, then $\bar{A} = A \cup \{a\} = (a,\infty)\cup\{a\}$ (I really don't get this. If the closure is a closed set, then its complement must be open, however $\bar{A}^c = (-\infty,a)$, which doesn't look like can be a union of elements of $B$).
I tried to understand this solution by Michael de Guzman. (Exercise 5A-3)
Please answer my questions above in parenthesis.
Thanks.
https://michaeltdeguzman.files.wordpress.com/2012/02/topohw32.pdf1
Best Answer
There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p \in A$. If $x \le p$, let $O$ be any basic set that contains $x$, so $O = (a,\infty)$ with $x \in (a,\infty)$, or $a < x$. As $a < x \le p$, $p \in O \cap A$ and so $x \in \overline{A}$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^\downarrow := \{x \in \mathbb{R}: \exists p \in A: x \le p\} \subseteq \overline{A}$.
Now if $x \notin A^\downarrow$, this means that $\forall p \in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p \in A$) $(p ,\infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x \notin \overline{A}$.
This shows that $A^\downarrow = \overline{A}$, for $A \subseteq \mathbb{R}$.
In particular $\overline{(a,\infty)} = \mathbb{R}$ for all $a \in \mathbb{R}$, so I don't agree with the linked solution that claims $\overline{(a, \infty)}$ just adds the point $a$ to $(a,\infty)$. All open subsets are "right facing" so it's quite clear that $[a,\infty) =\{a\} \cup (a,\infty)$ is not even closed in this topology (its complement is "left facing").