Proof the set of all interior points of a set E is open I am going over the Exercises in Baby Rudin and I have come across this one:
“Prove that the set of all interior points is always open”.I looked at the solution but it’s quite different from mine, so I would ask you to point out all the mistakes I am making.
So I would set up a proof in this way:Suppose that E is closed then $E^{\circ}$ which is the set of all interior points, is open.Conversely, if E is open we prove that $E^{\circ}$ is open anyway.
1)If E is open then  $E^{\circ}$ is open: I think this should follow by the definition of open set:every point of an open set is an interior point so we would have $E=E^{\circ}$ therefore  $E^{\circ}$ is open.
2)If E is closed then  $E^{\circ}$ is open:
So E is closed therefore we can have two cases:$E^{\circ}=E$ which is impossible because not every point of a closed set is an interior point, or $E^{\circ}\subset E$.But then this would mean that $E^{\circ c}$ is open and non empty so by setting some $\delta = \frac{d(p,q)}{2}$where $p \in E^{\circ c} $and $q \in E^{\circ }$we can find some point z which doesn’t lie in $E^{\circ}$ contradicting the assumption that all the interior points lie in $E^{\circ}$.
I don’t know it seems messy but I would like to hear your opinions and corrections.Thank you for all the help.
 A: *

*The property that you seem to be using everywhere:


Every set is either closed or open.

is very false. Consider the metric space $(\mathbb{R}, d)$ with the usual metric $d(x, y) = |x-y|$. Then:

*

*$(0, 1)$ is open but not closed

*$[0, 1]$ is closed but not open

*$\mathbb{R}$ is both closed and open (!)

*$[0, 1)$ is neither closed nor open.

So in general there is no logical relation between being an open set and being a closed set. As one said: sets are not doors. ;)


*The proof in the case $E$ is open is correct, although it doesn't give much - the argument has to be made in generality anyway.


*The proof in the case $E$ is closed is uncomprehensible: how do you know $E^{\circ c}$ is open and non-empty? How do you know there is any $q \in E^{\circ}$? How exactly do you find $z$ and why should it be an interior point?
Additional advice: what I think you should focus on is to very strictly stick to the definitions.

A set $G \subseteq X$ is open if for each $a \in G$ there is $r > 0$ such that $B(a, r) \subseteq G$.

So in order to prove that $E^{\circ}$ is open for an arbitrary set $E \subseteq X$, you should take any $a \in E^{\circ}$ and find $r > 0$ satisfying $B(a, r) \subseteq E^{\circ}$.
Also drawing a picture illustrating the problem for $X = \mathbb{R}^2$ might be very helpful.
