Let $A,B$ subsets of a normed space $(X,\|\cdot\|)$ and $A+B=\{a+b\mid a\in A,\, b\in B\}$
I need help with the next proofs, I can't figure how to begin the proofs:
(a) If $A,B$ open then $A+B$ open
(b) If $A=\{(x,y)\in\mathbb{R}^2 \mid x>0 \:\&\: xy\ge1\},\: Y=\{(0,y)\in\mathbb{R}^2 \mid y\in \mathbb{R}\}$ then $A,Y$ closed on $\mathbb{R}^2$ (Maybe prove it using sequences?)
(c)$A+Y=\{(x,y) + (0,z) \mid (x,y) \in A,\: (0,z)\in Y\} = \{(x,y)\in \mathbb{R}^2 \mid x>0\}$ is not closed
For the (a) question I was thinking if I can see it as $$\bigcup a+B \quad \forall a\in A$$ but I don't know how to prove $$a+B=\{a+b\mid b \in B\}$$ is an open set (if it's true it's an open set).
For the (c) question I should prove that $A+Y$ is open or I should prove that given a sequence in $A+Y$ it doesn't converge to a point of $A+Y$ ?
Here's how I would begin the proofs:
a) Let $w$ be an arbitrary point of $A+B$. Then $w$ has the form $a+b$ with $a$ in $A$ and $b$ in $B$. $A+B$ is open if each of its points has a neighbourhood contained entirely in $A+B$. We know that there should be neighbourhoods $N_a$ and $N_b$ of $a$ and $b$ contained entirely in $A$ and in $B$, respectively, because these two sets are open. Can these neighbourhoods be used to find a neighbourhood of $w$ contained in $A+B$?
b) I'll start with $A$. Let $\{(x_n, y_n)\}$ be a sequence of points in $A$, and further suppose that it converges to some point $p=(x,y)$ on the plane. Can we show that $p$ is contained in $A$ using what we know about the sequence? We know that $x_ny_n\ge1$ and that $x_n$ is positive for every $n$.
c) Some geometric intuition may point to an example of a sequence in the set with a limit outside of the set: is some "border" of the set not included in the set's definition? Try to describe a sequence of points with $x>0$ approaching one of those points in the border.