some statements on sum of two subsets of plane. open, closed etc . $W=\{(x,y)\in\mathbb{R}^2: x>0,y>0\}$
$X=\{(x,y)\in\mathbb{R}^2: x\in\mathbb{R},y=0\}$
$Y=\{(x,y)\in\mathbb{R}^2: xy=1\}$
$Z=\{(x,y)\in\mathbb{R}^2: |x|\le 1,|y|\le 1\}$


*

*$W+X$ is open


True as sum of an open set and a closed set is always open.(W open, X closed)


*$X+Y$ is closed


False, Sum of two closed set need not  be a closed set. But I am not able to figure out in this case what will be the summed up set.


*$Y+Z$ is closed


True as sum a closed and a compact(Z) is closed.
Thanks for helping and correcting me
 A: This is related to Is the sum of these two sets open or closed? 
But I will give answers here anyway (more questions here). 


*

*Correct, and a bit more is true: The sum of an open set and any set is open.   

*The sum is the plane, except the $x$-axis $\{(x,y):y\not=0\}$. Indeed if $(x_1,0)\in X$ and $(x_2,y_2)\in Y$ then clearly $y_2=\frac1{x_2}\not=0$, so 
$(x_1,0)+(x_2,y_2)=(x_1+x_2,y_2)$ cannot be on the $x$-axis.
Conversely, 
if $(x,y)$ with $y\not=0$ then $(x,y)=(x-\frac1y+\frac1y,y)= (x-\frac1y,0) + (\frac1y,y)\in X+Y$.
Alternatively think of the effect of adding $(x,0)$ to any set, this is a translation of the set along the horizontal axis by the vector $\langle x,0 \rangle$. So, $Y$ is the hyperbola, the graph of $xy=1$ (one branch in the I-st Quadrant, and one branch in the III-rd Quadrant). We translate $Y$ to the left and to the right, all possible ways, sweeping the upper and the lower half planes, except the points on the $x$-axis (since $Y$ has no points on the $x$-axis). Of course, removing the $x$-axis leaves an open (but not closed) set. 

*It is correct. 
