$\{ (x,y) \in \mathbb R^{2} | x>0, y \in \mathbb R \}$ not clopen? Let $S_{1} = \{ (x,y) \in \mathbb R^{2} | y \geq \frac{1}{x}, x> 0
\}$ and $S_{2} = \{ (x,y) \in \mathbb R^{2} | x = 0, y \leq 0 \}$. Now $S_{1} + S_{2} = \{ (x,y) \in \mathbb R^{2} | x > 0, y \in \mathbb R \}$, not clopen. Why is it not open? It certainly does not contain all boundary points such as $(0,0)$ but for open, every point requires a neighborhood (in Euclidean space, more here).
What is the $S_{1} + S_{2}$? It is not union but someting else? What about $S_{1} \cup S_{2}$? What is it like? It does not contain the boundary points so it is not closed. But I am now uncertain because the $S_{1}+S_{1}$ is not closed. My intuition falls here short.
 A: For subsets of $S_1$ and $S_2$ of $\mathbb{R}^n$, you can write $S_1+S_2$ as a union of translates. Namely,


*

*Given $X\subseteq \mathbb{R}^n$ and $\mathbf{a}\in\mathbb{R}^n$, 
$$X+\mathbf{a} = \{ \mathbf{x}+\mathbf{a}\mid \mathbf{x}\in X.\}$$
It is not hard to show that if $X$ is open, then so is $\mathbf{X}+\mathbf{a}$; and that if $X$ is closed then so is $X+\mathbf{a}$.

*Given subsets $X,Y\subseteq \mathbb{R}^n$, then
$$\begin{align*}
X+Y &= \{\mathbf{x}+\mathbf{y}\mid \mathbf{x}\in X,\mathbf{y}\in Y\}\\
&= \bigcup_{\mathbf{y}\in Y} (X+\mathbf{y})\\
&=\bigcup_{\mathbf{x}\in X} (Y+\mathbf{x}).
\end{align*}$$
So here, $S_1$ and $S_2$ are each closed; their sum is an infinite union of closed sets, and thus may or may not be closed a priori (it's not closed in fact). 
(An infinite union of closed sets can easily be open: take for example
$$\bigcup_{n=3}^{\infty} \left[\frac{1}{n},1-\frac{1}{n}\right] = (0,1),$$
which is an open set expressed as a countable union of closed intervals). 
However, $S_1+S_2$ in this case happens to be open. The reason it is not clopen is because it is not closed. 
As for $S_1\cup S_2$, it's a union of closed sets, so it is closed (which boundary points do you believe it does not contain?); but it is not clopen (because it is not open). The point $(1,1)$ is in $S_1$, but no neighborhood of $(1,1)$ is completely contained in $S_1\cup S_2$.
(In fact, $\mathbb{R}^n$ does not have any clopen sets except for $\emptyset$ and $\mathbb{R}^n$; that's because $\mathbb{R}^n$ is connected).
