I'm working in baby Rudin. I think understand the ideas of interior and limit points, open and closed sets.
We have the theorem that $E$ is closed iff $E^c$ is open.
This gives me some idea about the relationship between a set and its complement. However, I find myself stuck when it comes to determining whether the following statements are true, and, if so, how to prove them:
1. If $E$ is not open then it is closed, and vice-versa.
- If $E$ is not open and $E^c$ is not closed then $E$ is closed and $E^c$ is open.
Is this statement equivalent to saying that a set is closed if its complement is open?
- Proving the following (This one is a question from Rudin):
Let $E^o$ be the set of all interior points of a set $E$. Prove $E^o$ is always open.
(This is my first time using this markup language so if anyone can suggest improvements or good practices I'd be happy to adopt them).
Thanks
Best Answer
"Opposite" as a term will only mislead you. The entire space $\mathbb{R}$ is both open and closed, as is the empty set $\emptyset$; neither is the "opposite" of itself. The complement of an open set is closed, and vice versa.
If $E$ is not open then it is closed, and vice-versa.
This is false. Counterexample: $[0,1)$.
if $E$ is not open and $E^c$ is not closed then $E$ is closed and $E^c$ is open.
False. Saying that $E^c$ is not closed is equivalent to saying that $E$ is not open, so the hypothesis is just "$E$ is not open". From this, as noted in 1., it doesn't follow that $E$ is closed – as mentioned, they are not "opposites".
Let $E^o$ be the set of all interior points of a set $E$. Prove $E^o$ is always open.
$x$ is an interior point of $E$ iff some open neighborhood $U$ of $x$ is contained in $E$. So if $y \in E^o$, then $y \in U \subseteq E$ for some open neighborhood $U$ of a point $x \in E$. But then $y$ is by definition an interior point of $E$, as this same $U$ is an open neighborhood of $y$ contained in $E$.