Let $A,B\subseteq \mathbb{R}^n$, define $A+B=\{a+b : a\in A,b\in B\}$ Then which of the following is/are true?
$(1)$ If $A$ and $B$ are open, then $A+B$ is open.
$(2)$ If $A$ open and $B$ close, then $A+B$ closed.
$(3)$ If $A$ is closed and $B$ open, then $A+B$ is open.
$(4)$ If $A$ and $B$ are closed, then $A+B$ is closed.
My attempt
To prove $(1)$
Let $a+b \in A+B$. Since $A$ and $B$ are open , there are $\epsilon_1,\epsilon_2$ such that $B(a,\epsilon_1)\subseteq A$ and $B(b, \epsilon_2) \subseteq B$
Here $B(a,\epsilon_1)=\{x\in \mathbb{R}^n : ||x-a||\lt \epsilon_1\}$
Let $\epsilon=min \{\epsilon_1,\epsilon_2\}$
Let $z\in B(a+b,\epsilon) $ be arbitary.
Then $||(a+b)-z||\lt \epsilon$
or $||(a-(z-b)||\lt \epsilon \lt \epsilon_1$
$\Rightarrow z-b=a_1 \in A$
$\Rightarrow z=a_1+b \in A+B$
Hence $B(a+b,\epsilon) \subseteq A+B$ impling $A+B$ is open if $A$ and $B$ are open.
$(2)$ It is false by taking $A=(0,1) $ and $B=\{0\}$ in $\mathbb{R}$
To prove $(3)$
This is the kind of same as $(1)$
Let $a+b\in A+B$ .Since $B$ is open there is $\epsilon \gt 0$ such that $B(b,\epsilon)\subseteq B$
Let $z\in B(a+b,\epsilon)$
$||(a+b)-z||\lt \epsilon$
$\Rightarrow ||b-(z-a)||\lt \epsilon $
$z-a=b_1 \in B$
$z=a+b_1 \in A+B$ and thus $A+B$ is open.
$(4)$. It is false , I have two examples
Eg $(a)A=\mathbb{N}$ and $B=\{-n+\frac 1n: n\in \mathbb{N}\}$ in $\mathbb{R}$ .Then $A$ and $B$ being discrete subsets are closed. But $A+B$ cotains the sequence $\{1/n \}$ converging to $0$ but $0\notin A+B$
Eg $(b)$ A book I referred asked to think of $tanx$ on $(-\pi/2,\pi/2)$ in $\mathbb{R}^2$ but I can't proceed. Can you please help me complete it.?
Sorry for a long solution but being a budding mathematician, I want correct proofs . Can you please go through it and point out mistakes , if any?
Any alternative ideas will be appreciated. Thanks for your valuable time.
Your answers seem right to me.
Here is a more general way to look at (1) and (3).
Claim 1. If $A$ is open and $B$ is arbitrary then $A+B$ is open.
Proof. Note that $A+B=\bigcup_{b\in B}A+\{b\}$. Each set $A+\{b\}$ is open (by your argument for "open + closed is open"; but this specific case is easier.) Now $A+B$ is a union of open sets so it's open.
For $\tan x$ I guess you could do something like let $A$ be the graph of $\tan x$ on that interval and let $B=\{(0,r):r\in\mathbb{R}\}$. Those are closed sets but $A+B$ is $(-\pi/2,\pi/2)\times\mathbb{R}$ which is not closed. (I like your example better.)