Is if a set is not open and its complement not closed, is the set closed and its complement open?

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.

1. 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?

1. 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

• "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. Commented Oct 17, 2015 at 21:23
• Highly relevant: abstrusegoose.com/394 Commented Oct 17, 2015 at 21:24
• Hahaha. Thank you. Ok, I understand that 'opposite' is misleading. This answers 1). But 2) is still not clear to me. Commented Oct 17, 2015 at 21:29
• If $\mathbb R$ is our metric space, then $E = \mathbb Q$ is neither closed nor open, and $E^c = \mathbb R \setminus \mathbb Q$ is neither closed nor open. Commented Oct 17, 2015 at 21:32
• Take the interval $E:=[0,1)$ in $\mathbb{R}$, it is neither open nor closed. Moreover its complement is also not open and is not closed Commented Oct 17, 2015 at 21:34

"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.

1. If $$E$$ is not open then it is closed, and vice-versa.

This is false. Counterexample: $$[0,1)$$.

2. 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".

3. 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 $$E$$.

• But how does that let us to conclude $E^o$ is open? All we have specified is that each $y$ is indeed an interior point of $E$. Or is this basically a redundancy because all interior points of the set of interior points $E^o$ are interior points of $E$, so therefore, by definition of $E^o$, all such points are interior points of $E^o$, thus showing it is open? Commented Oct 17, 2015 at 21:53
• A set $X$ is open iff it contains a neighbhood of each of its points. And yes, the interior of the interior is the interior :) Does that clarify? Commented Oct 17, 2015 at 22:01
• Thank you! My comment was a bit convoluted... haha. Commented Oct 17, 2015 at 22:03
• Haha shared – not to worry, it takes time and dedication to get a firm grip on these ideas. Keep up the good work :) Commented Oct 17, 2015 at 22:20

In general a set may be neither open nor closed,and if this the case,then its complement is also neither closed nor open. The old joke is that "If X is not closed then X is open" is true if X is a door. As for Q.3. I don't know how Rudin defines $E^o$. Topologists define $E^o$ as the union of all open subsets of $E$ so of course it's open because the union of any family of open sets is an open set....... We may define $p\in E^o \iff \exists \text { open } V (p\in V\subset E).$ Now suppose $p\in E^o$. Choose an open set $V(p)$ where $p\in V(p)\subset E$. Observe that $\forall q\in V(p)$ we have [ $q\in V(p)\subset E \wedge V(p)$ is open.] So by the def'n of $E^o$,we have $\forall q\in V(p) (q\in E^o)$. That is, $V(p)\subset E^o$......... So if we let $F=\cup \{V(p) : p\in E^o\}$, we have $F\subset E^o$ . But $E^o=\{p :p\in E^o\}\subset \cup \{V(p) :p\in E^o\}=F.$ Therefore $F=E^o$.