Fix a point $b \in \mathbb R.$ The topology $\tau_{b}$ is generated by all intervals of the form $(x,\infty),$ where $x$ is less than $b$ as well as all intervals of the form $(-\infty,y),$ where $y$ is bigger than $b.$ In other words, the topology $\tau_b$ has subbasis given by
$$S = \{ (-\infty, y) \}_{y \in \mathbb R \,\,\text{&} \,\,y > b }\,\, \cup \{ (x, \infty) \}_{x \in \mathbb R \,\,\text{&} \,\,x < b }.$$ I want to find a countable basis for the topology $\tau_{b}$ , and determine if it is possible to find a metric on $\mathbb R$ whose metric topology is $\tau_{b}?$
$\mathbf{My\,\,attempt}:$
I'm a bit confused as to how $\tau_{b}$ differs from the standard topology on $\mathbb R.$ We know that the set of infinite open intervals $S' = \{ (-\infty, y) \}_{y \in \mathbb R}\,\, \cup \{ (x, \infty) \}_{x \in \mathbb R}$ is a subbasis for the standard topology on $\mathbb R.$ For $x < y, (-\infty, y) \,\,\cap \,\, (x,\infty) = (x,y)$ which implies that finite intersections of sets in $S'$ form a basis for standard topology. Hence, the collection of open intervals is a basis for the standard topology on $\mathbb R.$
I don't think fixing a point $b$ will change the answer. So am I right when I say that a countable basis for $\tau_{b}$ could be given by $\mathcal{B} = \{(x,y) \mid x,y \in \mathbb Q, x <y \}?$ If not, my vague guess is $\mathcal{B} = \{(x,y) \mid x,y \in \mathbb Q, x < b < y \}?$
Also, for the second-half of the question, will the usual metric on $\mathbb R$ defined by $d(x, y) = |x − y|$ work whose metric topology is $\tau_b?$
Best Answer
The thing you've missed is that every non-empty element of $\tau_b$ has $b$ as an element. The topology thus consists of $\emptyset,$ together with the open (not necessarily bounded) intervals around $b$ from the standard topology.
Your second attempt at a countable basis does the trick.
Do you see why this topology is not metrizable?