The way I have approached this question is this way:

  1. The set $\{1,2,3\}$ is not open because it does not contain any neighborhood at the point $x=1$
  2. 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.
  3. 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?

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    $\begingroup$ You're treating this set as a subset of what set, exactly? The reals? Integers? Rationals? $\endgroup$ – LuuBluum Oct 24 '16 at 19:27
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    $\begingroup$ Set with finite elements is closed $\endgroup$ – jnyan Oct 24 '16 at 19:28
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    $\begingroup$ @jnyan Not in every topology! :P $\endgroup$ – Noah Schweber Oct 24 '16 at 19:28

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?

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  • $\begingroup$ Ah, I see where I made the mistake then. The complement is open making the set closed. My logic was way off. How woudl you then find the boundary of the set {1, 2, 3} ∪ (2, 4)? $\endgroup$ – Aggrawal Puja Oct 24 '16 at 19:33

You don't need to show that every neighborhood of 0 is smaller than every $1/N$. You just need to find one neighborhood of 0 that is contained entirely within the complement. How about $(-1/2,1/2)$?

Because this can be done for any point in the complement, the complement is open, hence the set is closed.

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  • $\begingroup$ This makes sense, but what would the boundary for {1, 2, 3} ∪ (2, 4) be? $\endgroup$ – Aggrawal Puja Oct 24 '16 at 21:46

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