The boundary of $(-1,2)\cup[3,\infty)$ in $\mathbb{R}$ I am trying to determine the boundary of $S=(-1,2)\cup [3,\infty)$ within $\mathbb{R}$.
I thought that the set of boundary points would be all the points which contain an $\epsilon$-neighbourhood that has points in $S$ and points not in $S$. I thought that the only place that would be was $3$ because at $-1$ and $2$ you could have an $\epsilon$-neighbourhood small enough such that it would be fully contained in $S$ as these are limit points. Any explanation why the answer is given as $\{-1,2,3\}$?
 A: $-1$ is a boundary set because the following is true:

For every $\epsilon > 0$, the $\epsilon$ neighborhood of $-1$ includes both points in $S$ and points outside $S$.


Where you made your mistake is when you wrote the following:

at $-1$ and $2$ you could have an $\epsilon$-neighbourhood small enough such that it would be fully contained in $S$

This is not true.

More importantly, where you made your mistake is you wrote something without being certain if it is correct. In general, if you think $A$ is true because $B$ is true, you should always try to prove that $B$ is, in fact true. Without that proof, all you have is a hypothesis.
A: $S=(-1,2)\cup [3,\infty)$. Let $\partial S$ be the boundary of $S$. By definition:
$$\partial S = \overline{S} / int(S)$$
and
$$\overline{S} = [-1,2] \cup[3,\infty]$$
$$int(S) = (-1,2)\cup(3,\infty) \implies$$
$$\partial S = \overline{S} / int(S) = \{1,2,3\}$$
A: Doesn't matter how small ε-neighborhood of -1 you choose, you'll always get points from both inside and outside of S. You can divide the interval (-1-ε,-1+ε) into two parts, (-1-ε,-1) and (-1,-1+ε). All the points in (-1,-1+ε), belong to S, but every point of (-1-ε,-1) is outside S (regardless of choice of ε). So we can clearly see that -1 is a boundary point (even though it does not belong to the set S). With similar arguments you can also check that 2 and 3 are also boundary points. And hence, the boundary of S is {-1,2,3}. I hope it will clear your doubt:)
