# help with proof

• December 1st 2009, 04:13 PM
p00ndawg
help with proof
Prove the following:

An accumulation point of a set S is either an interior point of S or a boundary point of S.

This seems very logical to me, I just dont know how to prove it.
I know that an accumulation point may or may not be in S.
If it is in S then it would be an interior and boundary. If it is not in S then it would be a boundary point. Pretty simple I think, but how can I actually prove this?
• December 1st 2009, 04:57 PM
Drexel28
Quote:

Originally Posted by p00ndawg
Prove the following:

An accumulation point of a set S is either an interior point of S or a boundary point of S.

This seems very logical to me, I just dont know how to prove it.
I know that an accumulation point may or may not be in S.
If it is in S then it would be an interior and boundary. If it is not in S then it would be a boundary point. Pretty simple I think, but how can I actually prove this?

Let $E$ be a set and $\xi$ a limit point of $E$. Then for every neighborhood $N_{\varepsilon}(\xi)$ there contains a point of $E$ different from $E$. Note now that if we consider let $P$ be the statement "there exists some neighborhood of $\xi$ such that it is contained in $E$" then "Every neigborhood of $\xi$ contains a point of $E^{c}$" is the statement $\neg P$. And since we are working in first order logic it follows that $P\vee \neg P$ is a tautology. You get the idea.
• December 15th 2009, 07:44 PM
p00ndawg
I have been reading this post for awhile and the way you structured the proof does not make any sense to me.

Is there any other way to show this proof?
like with if X is in S or if x is not S sort of version then there exists a deleted neighborhood intersected with S that is nonempty?

I kind of have the jist of what accumulation points, boundary and interior points are but im having trouble forming a proof in my own words.
S is a subset of R.
I know that if x is an accumulation point there exists $N*(x, {\varepsilon} ) \cap$ S =/= $\emptyset$. So by that if x is in the deleted neighborhood and an accumulation point (meaning x is in S'), then x is not in S.
If x is not in S then x is in R/S.

So if x is in R/S then that means that $N(x , \varepsilon) \cap (R/S)$ =/= $\emptyset.$
So x in the bd of S.

Is this kind of on the right track?
• December 15th 2009, 08:25 PM
Drexel28
Quote:

Originally Posted by p00ndawg
I have been reading this post for awhile and the way you structured the proof does not make any sense to me.

Is there any other way to show this proof?
like with if X is in S or if x is not S sort of version then there exists a deleted neighborhood intersected with S that is nonempty?

I kind of have the jist of what accumulation points, boundary and interior points are but im having trouble forming a proof in my own words.
S is a subset of R.
I know that if x is an accumulation point there exists $N*(x, {\varepsilon} ) \cap$ S =/= $\emptyset$. So by that if x is in the deleted neighborhood and an accumulation point (meaning x is in S'), then x is not in S.
If x is not in S then x is in R/S.

So if x is in R/S then that means that $N(x , \varepsilon) \cap (R/S)$ =/= $\emptyset.$
So x in the bd of S.

Is this kind of on the right track?

I was a little lazy. I think if I rephrase my argument more rigorously it will make more sense.

Assume that $\xi\in S'$, then $\forall\varepsilon>0$ there exists some $x\in S,x\ne\xi$ such that $x\in N_{\varepsilon}(\xi)$. Now if there exists some $\varepsilon'>0$ such that $N_{\varepsilon'}(\xi)\subseteq S$ then $\xi\in S^{\circ}$. If not, then $\forall\varepsilon>0$ it is true that $N_{\varepsilon}(\xi)\nsubseteq S$ which means there exists some point $y\in S^{c}$ such that $y\in N_{\varepsilon}(\xi)$. Consequently, every neighborhood of $\xi$ contains a point of $S^c$, but since $\xi\in S'$ it is true that every neighborhood also contains a point of $S$. This tells us that $\xi\in\partial S$.

$S^{\circ}$-interior
$\partial S$-boundary
$S'$-limit points
$S^{c}$-complement
• December 15th 2009, 08:41 PM
p00ndawg
Quote:

Originally Posted by Drexel28
I was a little lazy. I think if I rephrase my argument more rigorously it will make more sense.

Assume that $\xi\in S'$, then $\forall\varepsilon>0$ there exists some $x\in S,x\ne\xi$ such that $x\in N_{\varepsilon}(\xi)$. Now if there exists some $\varepsilon'>0$ such that $N_{\varepsilon'}(\xi)\subseteq S$ then $\xi\in S^{\circ}$. If not, then $\forall\varepsilon>0$ it is true that $N_{\varepsilon}(\xi)\nsubseteq S$ which means there exists some point $y\in S^{c}$ such that $y\in N_{\varepsilon}(\xi)$. Consequently, every neighborhood of $\xi$ contains a point of $S^c$, but since $\xi\in S'$ it is true that every neighborhood also contains a point of $S$. This tells us that $\xi\in\partial S$.

$S^{\circ}$-interior
$\partial S$-boundary
$S'$-limit points
$S^{c}$-complement

Thank you, its kind of late so im really having trouble putting anything of value together but at first glance this makes ALOT more sense..

Thanks.
• December 16th 2009, 03:01 AM
HallsofIvy
I would be inclined to use indirect proof: Assume a point, p, is NOT either an interior point or a boundary point of S. Then p is an interior point of the complemennt of S. Prove that s cannot be an accumulation point of S.
• December 16th 2009, 07:39 AM
p00ndawg
Quote:

Originally Posted by HallsofIvy
I would be inclined to use indirect proof: Assume a point, p, is NOT either an interior point or a boundary point of S. Then p is an interior point of the complemennt of S. Prove that s cannot be an accumulation point of S.

the second part of the question says:
Prove the following:

(b)A boundary of a set S is either an accumulation point of S or an isolated point of S.

I tried doing it by contradiction but Im not sure you can do this part that way.
If I assume a point p is not an accumulation point nor an isolated point, then this point could be an interior point of S, or an interior point of R/S?
is that correct?
• December 16th 2009, 07:57 AM
Plato
Quote:

Originally Posted by p00ndawg
the second part of the question says:
Prove the following: (b)A boundary of a set S is either an accumulation point of S or an isolated point of S.

Suppose that $x\in\beta(S)$, the boundary.
If $x$ is an isolated point of $S$ we are done.
Suppose that it is not. For all open sets $\mathcal{O}$ if $x \in \mathcal{O} \Rightarrow \quad \left( {\exists y \in \mathcal{O} \cap S\backslash \{ x\} } \right)$.
What does that tell us?
• December 16th 2009, 08:11 AM
p00ndawg
Quote:

Originally Posted by Plato
Suppose that $x\in\beta(S)$, the boundary.
If $x$ is an isolated point of $S$ we are done.
Suppose that it is not. For all open sets $\mathcal{O}$ if $x \in \mathcal{O} \Rightarrow \quad \left( {\exists y \in \mathcal{O} \cap S\backslash \{ x\} } \right)$.
What does that tell us?

x and y are both in that intersection?
• December 16th 2009, 08:44 AM
Plato
Does it mean that every open set that contains $x$ contains a point of $S$ distinct from $x$?

What is that the definition of?
• December 16th 2009, 08:46 AM
p00ndawg
Quote:

Originally Posted by Plato
Does it mean that every open set that contains $x$ contains a point of $S$ distinct from $x$?

What is that the definition of?

accumulation point