Certain Subset of Sorgenfrey Plane is Closed

Question

Note that $L = \{(x,-x) \mid x \in \Bbb{R} \}$ is closed. Then if $A$ is closed in $L$, then it will also be closed in $\Bbb{R}^2_\ell$. According to Munkres, $L-A$ will also be closed, but I am having trouble proving this. The set $L-A$ closed in $\Bbb{R}^2_\ell$ if and only if $\Bbb{R}^2_\ell - (L-A) = (\Bbb{R}^2_\ell - L) \cup A$ is open, which I am having trouble seeing the truth of this.

Answer 1

As $\{(x,x)\} = ([x,x+1) \times [x,x+1)) \cap L$, every singleton subset of $L$ is open in $L$ (as an intersection of an open set of the Sorgenfrey plane with $L$).

This means that $L$ is discrete as a subspace: all of its subsets are open (and thus closed) in $L$ and as $L$ is closed in the Sorgenfrey plane, all of its subsets are closed in the Sorgenfrey plane.