What is the closure of a circle under the order topology? Find the closure of the set $A = \{ (x,y) \mid x^2 + y^2 = 1 \}$ under the topology with basis $B = \{ [a,b) \times [c,d) \subset \mathbb{R}^2 \mid a \lt b, c \lt d \}$
My working is as follows:
By definition, the closure of $A$ is define as the intersection of all closed sets containing $A$. All closed sets of $B$ are of the form $[a,b]\times [c,d]$ or $(a,b]\times [c,d]$ or $[a,b]\times (c,d]$ or $(a,b] \times (c,d]$. Hence the closure of $A$ must be $(-1,1] \times (-1,1]$. Have I interpreted this question correctly?
 A: It’s not true that all closed sets of $B$ are of the forms that you list. All of those sets are closed, but so are many others, including the set $A$, which is therefore its own closure.
There are several ways to see that $A$ is closed in $B$. One is to observe that every Euclidean open set in $\Bbb R^2$ is also open in $B$. Since closed sets are simply the complements of open sets, it follows immediately that every Euclidean closed set is closed in $B$ as well. Since $A$ is closed in the Euclidean topology on $\Bbb R^2$, it is therefore closed in $B$.
Alternatively, you can show directly that if $\langle x,y\rangle\in B\setminus A$, there is an $\epsilon>0$ such that
$$\big([x,x+\epsilon)\times[y,y+\epsilon)\big)\cap A=\varnothing\;;$$
then $[x,x+\epsilon)\times[y,y+\epsilon)$ is an open nbhd of $\langle x,y\rangle$ disjoint from $A$, so $\langle x,y\rangle\notin\operatorname{cl}A$.
By the way, the topology on $B$ is not any order topology. The most common name for the space $B$ is the Sorgenfrey plane.
