# Why is this last step even necessary in this proof with open sets?

Let $E^o$ denote the set of all interior point of a set $E$. Prove that $E^o$ is always open.

Proof: For $p \in E^o$, there is a neighborhood $N_r (p) \subset E$. Since neighborhoods are open, for $q \in N_r (p)$, there is a neighbourhood $N_{r'}(q) \subset N_r (p)$; $N_{r'}(q) \subset E$; $q in E^o$; $N_r(p) \subset E^o$; $E^o$ is open

Why is it necessary to show that a point inside the neighbor is interior and showing that as a subset of $E^o$ even necessary? Why can't we just say something like

Let $p_n \in E^o$, then $p_n$ is interior for each n. So there exists a neighborhood $N_r(p_n) \subset E^o$ for each $n$. So then by definition this set $E^o$ is open since it contains all its interior points $p_n$

EDIT:
I tried applying the proof to $E=[0,1]$ and I know that $E_0=(0,1)$. I decided to pick $p=0.5$ and $q=0.4$. Following the logic of the proof seems alright, but what concerns me is the last final touch: $q∈E_0;N_r(p)⊂E;q\in E$ is open How does that show E is open? Maybe it's because I drew picture but that step seem too obvious to "proof" anything –

eDIT

How also does $q \in E^0$ show that $N_r(p) \subset E^0$?

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This is because you must show that the neighborhood is in $E^o$. The neighborhood that you started off with is only in $E$. –  Calvin Lin Jan 1 '13 at 3:32
Why are we even picking points in $E$ to start with? What's wrong with picking points in $E^0$? –  sidht Jan 1 '13 at 4:12
You are given that $p$ is in the interior of $E$, which means that there is an open neighborhood in $E$, which contains $p$. You do not know that this open neighborhood is also in $E^0$, which is what the $N_{r'(q)}$ seeks to establish. –  Calvin Lin Jan 1 '13 at 4:53

## 2 Answers

Q 1.) Why can't we say the following?

Let $p_n\in E_0$, then $p_n$ is interior for each $n$.

Criticism: If the index $n$ is a natural number, then it seems you are implicitly assuming that number of interior points can be at most countable. That need not be true, consider $E=\mathbb{R}$ (the reals).

So there exists a neighborhood $N(p_n) \subset E_0$ for each $n$.

Criticism: From the definition of an interior point, you claim "$N(p_n) \subset \color{red}{E_0}$". The inference is incorrect. The right inference is "$N(p_n) \subset \color{blue}{E}$". Do you see how we end up back into the set $E$? Perhaps this should answer your next question...

Q 2.)"Why are we even picking points in $E$ to start with? What's wrong with picking points in $E_0$"

Ans: But points of $E_0$ are interior points of $E$!

To get an idea of what you are dealing with, try your proof on the following two examples:

1) $E = [0,1]$

2) $E = \{\frac1{n}:n \in \mathbb{N}\}$

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I tried applying the proof to $E = [0,1]$ and I know that $E^0 = (0,1)$. I decided to pick $p = 0.5$ and $q=0.4$. Following the logic of the proof seems alright, but what concerns me is the last final touch: $q \in E^0$;$N_r(p) \subset E^0$E is open How does that show E is open? Maybe it's because I drew picture but that step seem too obvious to "proof" anything –  sidht Jan 10 '13 at 6:16

Like Calvin points out above, an interior point of $E$ is not obviously also an interior point of $E^0$. The substance of the proof is taking a neighborhood around each interior point $p$ of $E$, concluding that all of the points in that neighborhood are also in $E^0$, and that therefore $p$ is an interior point of $E^0$.

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Why are we even picking points in E to start with? What's wrong with picking points in $E^o$ –  sidht Jan 1 '13 at 4:22
Sure, but picking points in $E^0$ is exactly picking interior points of $E$. –  user7530 Jan 1 '13 at 5:57