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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.

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    To answer the title, the set is open, but not closed (hence not clopen as well). This is because the the points $(0,y)$ are all limit points of the set, but they do not belong to the set. (But is this your question? Why do your title and body ask different questions?)2011-09-13
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    Did you mean to write "clopen"? A clopen set is a set that is *both* open and closed. Neither $S_1$ nor $S_2$ are are open.2011-09-13
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    @Arturo Magidin: yes, I meant clopen (open and closed). $S_{1}$ and $S_{2}$ are closed sets.2011-09-13
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    @hhh: Doesn't seem like there is a lot of connection with the Wikipedia quote, but okay.2011-09-13
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    @Arturo magidin: I always forget definitions so I included them there but I removed at your notice.2011-09-13
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    @hhh: $S_1+S_2$ **is** open; why is it you believe it is not open? What it is not is *closed*. It's not closed, but it doesn't have to be closed because, even though $S_1$ and $S_2$ are each of them closed, their *sum* is actually an infinite union of closed sets, and an infinite union of closed sets need not be closed.2011-09-13

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