Translate of a closed set is closed(part2)

Previously, I raised a question whether $$(a+F)^c=a+F^c.$$ Jonas Meyer pointed out that it is true. After which, I was able to prove the first inclusion. The details are as follows: let $y\in a+F^c$. Then there exists $z\in F^c$ such that $y=a+z$. We claim that $y\in(a+F)^c$. Suppose $y\notin (a+F)^c$. Then $y\in a+F$. Thus, there exists $x\in F$ such that $y=a+x$. This implies that $z=x$. Hence, $x\in F\cap F^c=\varnothing$. We obtain a contradiction. Thus, $y\in(a+F)^c$. Hence, $a+F^c \subseteq (a+F)^c$. I tried the other inclusion, but can't prove it. A help on this is very much appreciated.

juniven

-

By your first step $(a+ F)^c = a + (-a) + (a+F)^c \subseteq a + \bigl((-a)+ a + F\bigr)^c = a + F^c.$