# Thread: Proof check Accumulation point

1. ## Proof check Accumulation point

Suppose b = Sup S and b is not in S. prove b is an accumulation point.

if b is the supremum of S but not in S then b is in R/S.
Since b is the supremum then for all s $\displaystyle \in$ S, s < b.
But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c $\displaystyle \in$ S, but still less than b.

is this ok?

2. Originally Posted by p00ndawg
Suppose b = Sup S and b is not in S. prove b is an accumulation point.

if b is the supremum of S but not in S then b is in R/S.
Since b is the supremum then for all s $\displaystyle \in$ S, s < b.
But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c $\displaystyle \in$ S, but still less than b.

is this ok?
I don't understand why $\displaystyle b\notin S$ is needed?

Theorem: Let $\displaystyle E\subseteq R$ such that $\displaystyle \sup\text{ }E$ exists. Then $\displaystyle \sup\text{ }E$ is a limit point of $\displaystyle E$.

Proof: Suppose not. Then there exists a $\displaystyle \delta>0$ such that $\displaystyle N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing$, but by definition this means that there does not exists an $\displaystyle x\in E$ such that $\displaystyle \sup\text{ }E-\delta<x$. This is a contradiction.

3. Originally Posted by p00ndawg
Suppose b = Sup S and b is not in S. prove b is an accumulation point.

if b is the supremum of S but not in S then b is in R/S.
Since b is the supremum then for all s $\displaystyle \in$ S, s < b.
But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c $\displaystyle \in$ S, but still less than b.

is this ok?
.

Since b=SupS we have by definition that:

1) for all ,x : $\displaystyle x\in S\Longrightarrow x\leq b$

2) for all,ε>0 ,there exists a y belonging to S ,such that :

$\displaystyle b-\epsilon<y\leq b\Longrightarrow b-\epsilon<y<b+\epsilon$$\displaystyle \Longrightarrow |y-b|<\epsilon$ .................................................. .................................................. ..1

And since b does not belong to S we have :

$\displaystyle y\neq b$.................................................. .......................................2

Hence from (1) and (2) we conclude ,that there exists a deleted neighbourhood round b containing yεS.

So b is an accumulation point.

4. Originally Posted by Drexel28
I don't understand why $\displaystyle b\notin S$ is needed?

Theorem: Let $\displaystyle E\subseteq R$ such that $\displaystyle \sup\text{ }E$ exists. Then $\displaystyle \sup\text{ }E$ is a limit point of $\displaystyle E$.

Proof: Suppose not. Then there exists a $\displaystyle \delta>0$ such that $\displaystyle N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing$, but by definition this means that there does not exists an $\displaystyle x\in E$ such that $\displaystyle \sup\text{ }E-\delta<x$. This is a contradiction.
Drexel28 ,your theorem and your proof are both wrong.

Your theorem should be:

Theorem: Let $\displaystyle E\subseteq R$ such that $\displaystyle \sup\text{ }E$ exists .And SupE does not belong to E . Then $\displaystyle \sup\text{ }E$ is a limit point of $\displaystyle E$

And in your proof $\displaystyle N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing$ ,does not imply :

that there does not exists an $\displaystyle x\in E$ such that $\displaystyle \sup\text{ }E-\delta<x$.

But that:

for all ,x : $\displaystyle \neg (x\in E)$ or (SupE=x) or $\displaystyle \neg |x-SupE|<\delta$.

Which is equivalent to:

$\displaystyle SupE\neq x\Longrightarrow\neg(x\in E\wedge |x-SupE|<\delta)$.

And that shows that:

Only if $\displaystyle SupE\neq x$, then there does not exists an $\displaystyle x\in E$ such that $\displaystyle \sup\text{ }E-\delta<x$.

And since SupE does not belong to E we have:

$\displaystyle SupE\neq x$ and the contradiction follows.

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### proof on accumulation point

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