# Minkowski sum of open and closed set

Let $$A$$ and $$B$$ nonempty subsets of $$\mathbb{R}$$ such that $$A$$ is open and $$B$$ is closed. Then:

a) $$A+B$$ is open,

b) $$A+B$$ is closed,

c) $$A+ int B$$ is open,

d) $$A$$ difference $$B$$ is open.

I can prove option a) i. e. U (A+b) where b varies over the set B. For option b) I have the counter example $$A=(0, 1),B=[1, 2]$$ then $$A+B=(1, 3)$$.... For option c) $$A$$ is open and $$int B$$ is open hence their minkowski sum is open, but I am not getting option d

• i tried to suggest an edit but i think messed up some of the formatting. (In your counterexample A+B should be (1,3)
– A. P
Aug 18, 2019 at 13:31
• I am not sure about the exact range you wonder about, so I will try to write the full solution for you. Aug 18, 2019 at 13:45
• @A.P I have written A+B =(1, 3)...I am not getting your point. Aug 18, 2019 at 20:01
• np, I extend my answer
– A. P
Aug 18, 2019 at 20:40

$$A - B$$ is defined as $$\{c| c+B \subset A\}$$ this is equivalent to: $$A-B=(A^c+(-B))^c$$ (I define $$(-B)$$ as $$\{-b|b \in B\}$$)
The equivalence holds since $$(A-B)^c=\{c| c+B\not\subset A\}=\{c|\exists b \in N \text{ with } c+b \in A^c\}= \{A^c-b| b\in B\}=\{A^c+b| b\in (-B)\}=A^c +(-B)$$
Since $$B$$ and $$A^c$$ are closed it follows that $$A^c +(-B)$$ is closed. Therefore $$(A^c +( -B))^c = A-B$$ is open.