Proof that p is interior if p is not limit of complement

[jamalkoiyess]

Hello PF,

I am searching for a proof that I couldn't find on the internet.

Theorem: E in X a metric space. p in E. p is an interior point of E if and only if p is not a limit point of (E complement)'

Sorry for notations but I have no idea how to insert Latex here.

 

[member 587159]

Hello PF,

I am searching for a proof that I couldn't find on the internet.

Theorem: E in X a metric space. p in E. p is an interior point of E if and only if p is not a limit point of (E complement)'

Sorry for notations but I have no idea how to insert Latex here.

What have you tried? Contradiction seems a promising method. (Didn't try yet)

 

[jamalkoiyess]

What have you tried? Contradiction seems a promising method. (Didn't try yet)

I didn't even try...
I will see if contradiction works.

 

[member 587159]

I didn't even try...
I will see if contradiction works.

Well, it are the forum rules that you at least try yourself, and post what you tried.

 

[jamalkoiyess]

Well, it are the forum rules that you at least try yourself, and post what you tried.

So this is the proof I came up with, tell me if something is wrong:
$E \in X$
Suppose that $p \in \mathring{E}$ and $p \in (E^c)'$
Then $\exists \quad \gamma>0 \quad s.t. \quad B_\gamma(p) \in E$
and $\forall \quad r>0 \quad \exists q \in E^c \cap B_r(p)$
now let $r = \gamma$
So $q \in B_r(p) \subset E \quad and \quad q \in E^c$
contradiction.

 

[member 587159]

So this is the proof I came up with, tell me if something is wrong:
$E \in X$
Suppose that $p \in \mathring{E}$ and $p \in (E^c)'$
Then $\exists \quad \gamma>0 \quad s.t. \quad B_\gamma(p) \in E$
and $\forall \quad r>0 \quad \exists q \in E^c \cap B_r(p)$
now let $r = \gamma$
So $q \in B_r(p) \subset E \quad and \quad q \in E^c$
contradiction.

Yes, one small mistake $B(p) \subseteq E$ instead of $B(p) \in E$. And you showed only one implication ( the other one is pretty much the same I think though)