The way I have approached this question is this way:
- The set $\{1,2,3\}$ is not open because it does not contain any neighborhood at the point $x=1$
- The complement of the set contains $0$. However, if $(-a, a)$ is any neighborhood of $0$, then there exists an $N$ so large such that $1/N < a$. Because this neighborhood is not part of the complement, it contains the element $1/N$ from the set. Therefore the complement is not open.
- This has shown that the set is neither open or closed.
Is this approach right?
What I then struggle with is finding the boundary of the set ${1, 2, 3} \cup (2, 4)$? If the first part is correct, how would I find this boundary?
Best Answer
Presumably you're working inside $\mathbb{R}$ with the usual topology? This has to be made explicit - change the ambient topology, and the answer can change!
Your (1) is correct - the set $\{1, 2, 3\}$ is not open. However, I do not understand your (2): if $a$ is small enough, then $(-a, a)$ is an open interval containing $0$ disjoint from $\{1, 2, 3\}$ - how does that imply that $\{1, 2, 3\}$'s complement is not open? Remember that the complement of $A$ is open if for all $x\not\in A$, there is an open $U\ni x$ with $U\cap A=\emptyset$.
If you believe that the complement of $\{1, 2, 3\}$ is not open, here's what you need to do:
Pick some $x\not\in \{1, 2, 3\}$.
Show that any open $U\ni x$ is not disjoint from $\{1, 2, 3\}$ (that is, $U$ contains $1$, or $U$ contains $2$, or $U$ contains $3$).
You've picked $x=0$ - but you haven't correctly argued that any open set containing $0$ intersects $\{1, 2, 3\}$. Do you understand why?