# topology for engineers: closed, open, and bounded sets

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?

• 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. – Jendrik Stelzner Jan 9 '16 at 16:18
• 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. – Gyro Gearloose Jan 9 '16 at 17:03
• And, a bounded set need neither be open nor closed. Think of $(0,1]$. And there are far worse examples. – Gyro Gearloose Jan 9 '16 at 17:09
• wait, Gyro Gearloose, is that example not just closed? It includes a boundary, so that would contradict my first statement. – science404 Jan 9 '16 at 17:26
• 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. – Gyro Gearloose Jan 9 '16 at 17:56

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 $$S \subset \overline{B}(a,R) \Rightarrow S \subset B(a,R+1)$$