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.

  • $\begingroup$ This are the exact words in the book "A point $z_0$ is said to be an interior point of a set $S$ whenever there is some neighborhood of $z_0$ that contains only points of $S$". Is it different from what I wrote? Or is it different from what you wrote? They look pretty similar to me, please tell me if there's some gap between them $\endgroup$ Dec 3, 2016 at 18:48
  • $\begingroup$ Forgive me, I misunderstood you definition. However it is a bit confusing because it seems that you are referring to a specific point $z$. Specifically, if I say that $A$ contains only $z\in S$ what I immediately think is that $A=\{z\}$, whatever $z$ is, while you only meant $A\subseteq S$. $\endgroup$
    – Caligula
    Dec 3, 2016 at 18:54
  • $\begingroup$ I removed the "logic" tag - that's for questions within the specific field of mathematical logic. $\endgroup$ Dec 3, 2016 at 19:00

1 Answer 1


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 $U\subseteq S$.

  • $\begingroup$ I like your wordy proof, I'm quite a fan of these, as in most of my homework I exactly do them. However I wanted to train myself in proving things in a "tedious" way, i.e. using a logic definition and using either negations of it or contrapositives to get to the statement we want to prove. Would you know how I could rephrase those definitions or actually going from there to the wanted result? Thanks $\endgroup$ Dec 3, 2016 at 18:58
  • $\begingroup$ I edited my answer with some more logic inside it. $\endgroup$
    – Caligula
    Dec 3, 2016 at 22:12
  • $\begingroup$ Thank you! It makes sense! It's a nice proof, based a lot on sets! I have no idea what a filter is in mathematics, however the set $U_z$ makes sense $\endgroup$ Dec 4, 2016 at 2:13

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