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Let $E^o$ be the set of all interior points of the set $E$.

I was able to prove that $E^o$ is always open and $E$ is open iff $E = E^o$.

Now I am asked to prove that $(E^o)^c = \overline {E^c}$.

Intuitively it is very clear, but I am not sure if my proof is valid.

proof:

Let $x$ be a point not in $E^o$. Then either $x$ is a limit point of $E$ ($x\in {E^c}'$) or not in $E$ ($x \in E^c$). This means that all x-neighborhoods $V_x$ is not a subset of $E^o$, so there is always a point $y$ out side $E^o$,different from $x$, inside $V_x$. ($\exists y \notin E^o \text{ s.t } y \ne x , y \in V_x$). This means that $x \in E^c \cap {E^c}'= \overline {E^c} $ and $\overline {E^c}$ is closed. Thus $(E^o)^c = \overline {E^c}$.

The following is what I would really like to know.

a), How is $E^o$ read?

b), The over all idea is to show that if $x$ is not in $E^o$, then it must be in the closure of the complement of $E$. Am I right ?

c), If my proof is not valid, could you show me an example of a proof ?

I am teaching myself analysis and never took topology. My knowledge is limited up to abstract algebra, so please keep that in mind. Thank you.

hyg17
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1 Answers1

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a) I always read it as "the interior of $E$". I believe it is the most common term.

b) That would be the standard way of attacking this problem. You're also right about separating the two cases of $x$ being a limit point of or contained in $E^c$.

However, the proof is not quite complete yet. You've shown that $(E^o)^c \subseteq \overline {E^c}$. To complete the proof, you need to show that $(E^o)^c \supseteq \overline {E^c}$, that is, take a point $y \in \overline {E^c}$, and show that $y\in (E^o)^c$.

Arthur
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