I am working through the problems in Topology by Munkres. This comes from Section 17, #17 on page 101.
Consider the lower limit topology on $\mathbb{R}$ and the topology given by the basis $\mathcal{C}$, where $\mathcal{C} = \{[a,b)\text{ such that }a \lt b, a\text{ and }b \text{ rational}\}$. Determine the closure of the intervals $A = (0, \sqrt{2})$ and $B = (\sqrt{2}, 3)$ in these two topologies."
Here is my answer.
Closure of $A = (0, \sqrt{2})$ in the lower limit topology is $[0, \sqrt {2})$. Every neighborhood around $0$ must intersect $A$, so $0$ is an element of $\overline{A}$. Also, $[\sqrt 2, 4)$ is an open set which doesn't intersect $A$, thus $\sqrt{2}$ is not an element of $\overline{A}$.
Closure of $B = (\sqrt{2}, 3)$ in the lower limit topology is $[\sqrt{2}, 3)$. Every neighborhood around $\sqrt{2}$ must intersect $B$, so $\sqrt{2}$ is an element of $B$-closure. Also, $[3, 4)$ is an open set which doesn't intersect $B$, thus $3$ is not an element of $B$-closure.
Closure of $A = (0, \sqrt{2})$ in the topology generated by the basis $\mathcal{C}$ is $[0,\sqrt{2}]$. Every neighborhood around $0$ must intersect $A$, so $0$ is an element of $A$-closure. There do not exist open sets like $[\sqrt{2}, 4)$, where $\sqrt{2}$ is the lowest element in the set. Thus, any open set containing $\sqrt{2}$ must contain an element lower than $\sqrt{2}$. Thus $\sqrt{2}$ is an element of $A$-closure.
Closure of $B = (\sqrt{2}, 3)$ in the topology generated by the basis $\mathcal{C}$ is $[\sqrt{2}, 3)$. Every neighborhood around $\sqrt{2}$ must intersect $B$, so $\sqrt{2}$ is an element of $B$-closure. Also, $[3, 4)$ is an open set which doesn't intersect $B$, thus $3$ is not an element of $B$-closure.
Are these correct? Any feedback would be most appreciated.
Thanks.
Best Answer
These are all correct. It's worth noting, though, that you should justify (for example) why there are no other points in the closure of $(0,\sqrt2)$ in the lower limit topology. You've demonstrated why $\sqrt2$ isn't in the closure, but why isn't there any element greater than $\sqrt2$? Why isn't there any negative element?