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

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    $\begingroup$ i tried to suggest an edit but i think messed up some of the formatting. (In your counterexample A+B should be (1,3) $\endgroup$ – A. P Aug 18 '19 at 13:31
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    $\begingroup$ I am not sure about the exact range you wonder about, so I will try to write the full solution for you. $\endgroup$ – C.Park Aug 18 '19 at 13:45
  • $\begingroup$ @A.P I have written A+B =(1, 3)...I am not getting your point. $\endgroup$ – Soumyadweep Mondal Aug 18 '19 at 20:01
  • $\begingroup$ np, I extend my answer $\endgroup$ – A. P Aug 18 '19 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.

  • $\begingroup$ Edit summary: I changed my answer form a hint to an answer $\endgroup$ – A. P Aug 18 '19 at 20:53

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