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I have a question, from an engineers point of view (my apologies, rough on the math). From what I have gathered from questions on topology here, considering sets in $\mathbb{R}^d$:

  • a closed set has (and includes) a boundary
  • an open set may (or may not) have a boundary. If so, doesn't include that boundary
  • a bounded domain fits in a ball, and it could be open or closed

Are these dumbed down explanations close to correct?

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  • $\begingroup$ Roughly speaking this is right. More precisely, one can formally define what the boundary of an arbitrary subset $S \subseteq \mathbb{R}^d$ is. $S$ is then closed if and only if it includes its boundary and open if and only if it does contains no points of its boundary. $\endgroup$ – Jendrik Stelzner Jan 9 '16 at 16:18
  • $\begingroup$ And bear in mind that there are even sets that are open and closed. $\emptyset$ and $\mathbb{R}^d$ for one, but if you are looking at $\mathbb R$ with holes, there are more examples. Think of where $\frac 1{x^2-1}$ is defined. Excluding $\pm 1$ makes $(-1,+1)$ open and closed. $\endgroup$ – Gyro Gearloose Jan 9 '16 at 17:03
  • $\begingroup$ And, a bounded set need neither be open nor closed. Think of $(0,1]$. And there are far worse examples. $\endgroup$ – Gyro Gearloose Jan 9 '16 at 17:09
  • $\begingroup$ wait, Gyro Gearloose, is that example not just closed? It includes a boundary, so that would contradict my first statement. $\endgroup$ – science404 Jan 9 '16 at 17:26
  • $\begingroup$ For $\mathbb R\setminus \{-1,+1\}$ $(-1,+1)=\bigcup_{1>\epsilon>0}(-1+\epsilon,+1-\epsilon)$. As arbitrary unions of open sets are open, this is open. $\endgroup$ – Gyro Gearloose Jan 9 '16 at 17:56
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A subset of $\mathbb{R}^d$ always has a boundary which might be empty.

The boundary of a closed set can be empty. In particular, this is the case for $\mathbb{R}^d$ itself.

And you're right a bounded domain $S$ is included in a ball (by definition). Such a ball can be chosen open or closed as $$S \subset B(a,R) \Rightarrow S \subset \overline{B}(a,R)$$ and $$S \subset \overline{B}(a,R) \Rightarrow S \subset B(a,R+1)$$

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  • $\begingroup$ ok i didn't think of an empty boundary. that is a useful detail. $\endgroup$ – science404 Jan 9 '16 at 16:44

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