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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

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    $\begingroup$ "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. $\endgroup$ – BrianO Oct 17 '15 at 21:23
  • $\begingroup$ Highly relevant: abstrusegoose.com/394 $\endgroup$ – PhoemueX Oct 17 '15 at 21:24
  • $\begingroup$ Hahaha. Thank you. Ok, I understand that 'opposite' is misleading. This answers 1). But 2) is still not clear to me. $\endgroup$ – socrates Oct 17 '15 at 21:29
  • $\begingroup$ 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. $\endgroup$ – user217285 Oct 17 '15 at 21:32
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    $\begingroup$ 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 $\endgroup$ – Maxime Scott Oct 17 '15 at 21:34
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"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 $E$.

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  • $\begingroup$ 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? $\endgroup$ – socrates Oct 17 '15 at 21:53
  • $\begingroup$ 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? $\endgroup$ – BrianO Oct 17 '15 at 22:01
  • $\begingroup$ Thank you! My comment was a bit convoluted... haha. $\endgroup$ – socrates Oct 17 '15 at 22:03
  • $\begingroup$ Haha shared – not to worry, it takes time and dedication to get a firm grip on these ideas. Keep up the good work :) $\endgroup$ – BrianO Oct 17 '15 at 22:20
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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 $F=E^o$.

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