Q:

Precise proof of S Open set iff contains only interior points - Complex Analysis

In "Complex Variables and Applications" by Brown and Churchill (McGraw-Hill) the proof of: $$\textit{S is an open set $\implies$ each of its points is an interior point}$$ is left as an exercise. Here are the important definitions that one can use: $$\textbf{$\epsilon$ Neighbourhood}: |z-z_0|<\epsilon$$ $$\textbf{$z_0$ Interior point}: \exists\,\,\ \text{an}\,\,\, \epsilon\text{-neighbourhood containing only}\, z\in S$$ $$\textbf{$z_0$ Exterior point}: \exists\,\,\ \text{an}\,\,\, \epsilon\text{-neighbourhood containing no}\, z\in S$$ $$\textbf{$z_0$ Boundary point}: \text{all neighbourhood of $z_0$ have at least a point in $S$ and one not in $S$}$$ $$\textbf{$S$ Open}: \text{does not contain any of its boundary points}$$
I've tried to prove this in the following way, however I got stuck. Can someone tell me how to make it mathematically rigorous, using the above definitions?
$S$ is an Open Set $\implies$ $S$ does not contain any any of its boundary points.
So $$\text{if} \,\, [\forall \epsilon>0, (\exists z_1\in S\,\,\ \text{and}\,\,\ \exists z_2\notin S ):(|z_1-z_0|<\epsilon \,\,\text{and} \,\,\ |z_2-z_0|<\epsilon)] \implies z_0\notin S$$
Hence  taking the converse of the above we have $$\text{if}\,\, z_0\in S \implies \exists \epsilon>0, (\forall z_1\in S\,\,\ \text{or}\,\,\ \forall z_2\notin S ):|z_1-z_0|\geq \epsilon \,\,\text{or} \,\,\ |z_2-z_0|\geq \epsilon)$$
Which doesn't really tell me much.
I am really not sure I've even negated it correctly or that my definition with the quantifiers is correct.

A:

Using your definitions (of which some are not very well stated), we can proceed this way.
You want to prove that each point $z$ of an open set $S$ is an interior point. Now, since $S$ is open, $z$ can't be a boundary point, so negating the definition of boundary point, it follows that there exists a neighborhood $U$ of $z$ such that two things can happen:

or $U$ contains only point of $S$;
or $U$ contains only point outside $S$.

Since $U$ intersect $S$ at least in $z$, the latter is impossible, so you have proved that $U\subseteq S$. This means that $z$ is interior in $S$.

Following the comment, I'll try to rephrase the proof in more logically strict terms - but note that using words instead of quantifiers is not something to despise, as it increases greatly the clearity of proofs.
The fact $S$ is open means that

$\forall\,z\in S, z\notin \partial S$

where $\partial S$ is the boundary. Now, $z\in \partial S$ if and only if

$\forall \,U\in\mathscr{U}_z,\,U\cap S\neq \varnothing \text{ and } U\cap (S^c)\neq \varnothing$

where $(-)^c$ is the complement and $\mathscr{U}_z$ denotes the set of all open neighborhoods for $z$ (it has the structure of a filter if you know what it means). Hence, negating the latter will imply that

$\exists \,U\in \mathscr{U}_z$ such that $U\cap S = \varnothing$ or $U\cap (S^c)=\varnothing$

Then you conclude as before: the only possible conclusion is that $U\cap (S^c)=\varnothing$ and that is equivalent to $U\subseteq S$.

