Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $N_r(p) \subset E^0$?

share|improve this question
1  
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 $E^0$? –  jip 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 2

up vote 1 down vote accepted

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

share|improve this answer
    
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 –  jip 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 $E^0$.

share|improve this answer
    
Why are we even picking points in E to start with? What's wrong with picking points in $E^o$ –  jip 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.