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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?) –  Srivatsan Sep 13 '11 at 19:11
    
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. –  Arturo Magidin Sep 13 '11 at 19:11
    
@Arturo Magidin: yes, I meant clopen (open and closed). $S_{1}$ and $S_{2}$ are closed sets. –  hhh Sep 13 '11 at 19:14
    
@hhh: Doesn't seem like there is a lot of connection with the Wikipedia quote, but okay. –  Arturo Magidin Sep 13 '11 at 19:17
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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. –  Arturo Magidin Sep 13 '11 at 19:27
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1 Answer

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,

  1. 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 $X+\mathbf{a}$.

  2. 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).

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because "$\mathbb R^{n}$ is connected"? –  hhh Sep 13 '11 at 19:40
    
Wow, very good info here about translations. Yes, it is a translation, good formulation. Good, good. +1 –  hhh Sep 13 '11 at 19:44
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@hhh: A topological space $X$ is "disconnected" if and only if there exist nonempty open sets $A$ and $B$ such that $A\cap B=\emptyset$ and $A\cup B=X$; it is "connected" if it is not disconnected. It an easy consequence of the definitions that $X$ is disconnected if and only if there is a nonempty proper subset $A$ of $X$ that is clopen, and that $X$ is connected if and only if the only clopen subsets are $\emptyset$ and $X$. $\mathbb{R}^n$ is connected for every $n$. –  Arturo Magidin Sep 13 '11 at 19:46
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