Prove that x is in S or x is an accumulation point of S

[Eclair_de_XII]

Homework Statement

"Let $S$ be a non-empty set of real numbers bounded from above, and let $x=sup(S)$. Prove that $x \in S$ or $x$ is an accumulation point of S."

Homework Equations

Neighborhood: "A subset $O$ of $ℝ$ is a neighborhood of $x\in ℝ$ iff there exists an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon) \subseteq O$."
Accumulation point: "A number $x \in ℝ$ is an accumulation of a set $S$ if any neighborhood of $x$ contains an infinite number of elements in $S$."

The Attempt at a Solution

"Since $S$ is a non-empty set of real numbers bounded from above, there exists a least-upper bound of $S$, by the least-upper-bound property of $ℝ$. Let $x=sup(S)$. Then we wish to prove that $x\in S$ or $|(x-\epsilon,x+\epsilon) \cap S|=\infty$.

First, we note that the interval, $(x,x+\epsilon)\cap S$ is empty, because there exist no $a\in S$ such that $a > x=sup(S)$. Thus, we need to consider the cardinality of the set $(x-\epsilon,x]\cap S$.

We start by observing that $x-\epsilon$ is not an upper bound for $S$, since $x-\epsilon<sup(S)$. Hence, there exists an $a \in S$ such that $a>x-\epsilon$. Consequently, since $x$ is an upper bound of $S$, $x \geq a > x-\epsilon$. Thus, the interval $(x-\epsilon,x]\cap S$ is non-empty.

If $S$ was a finite set of real numbers, then it has a most element. And that most element was $x=sup(S)=max(S)\in S$."

Then I wanted to claim that if $S$ was infinite, then $|(x-\epsilon,x] \cap S|=\infty$. But then I thought of a set where $S$ could be infinite, and yet the latter statement about its intersection with $(x-\epsilon,x]$ was untrue.

Suppose $S=\{x-\frac{\epsilon}{2}\}\cup (-\infty,x-\epsilon]$. Then $|(x-\epsilon,x] \cap S|=1$, but $S$ would still be an infinite set, and $x$ still couldn't be an accumulation point of $S$. Additionally, I'm wondering why $x$ can't be in $S$ and be an accumulation point for $S$ at the same time. For example, the interval $[0,1]$ contains the element $1$, which is also its supremum, and an accumulation point for the interval.

Can anyone tell me any flaws or leaps in logic I may have made? Thank you...

 

[haruspex]

then I thought of a set where S could be infinite, and yet the latter statement about its intersection with (x−ϵ,x] was untrue

Quite so. S being infinite doesn't help. Its intersection with (x−ϵ,x] could be finite.
Can you construct an infinite sequence of points in S converging to x?

why x can't be in S and be an accumulation point for S at the same time.

That is allowed. The "or" in the condition to be proved is inclusive.

 

[Eclair_de_XII]

Can you construct an infinite sequence of points in S converging to x?

I suppose I can: $\{a_n\in S: a_n = x-\frac{\epsilon}{n},2\leq n\in ℕ\}$.

 

[haruspex]

I suppose I can: $\{a_n\in S: a_n = x-\frac{\epsilon}{n},2\leq n\in ℕ\}$.

I don't think that works. You are choosing some fixed ε. You have no guarantee that any of the x-ε/n are in S.

 

[Eclair_de_XII]

Let me see if I have this straight:

For my proof to work, I must devise a sequence that is a subset of $S$ that converges to the accumulation point for all $\epsilon>0$, since for $x$ to be an accumulation point, every neighborhood of $x$ must contain infinitely many points of $S$. So it isn't necessary that $(x-\epsilon,x] \subseteq S$. So it's possible that $x-\epsilon<inf(S)$ and that $x-\frac{\epsilon}{n} \notin S$.

Also, even if I have this sequence, how can I guarantee that it exists in $S$? The problem states only that $S$ is non-empty, not infinite.

 

[haruspex]

a sequence that is a subset of S that converges to the accumulation point for all ϵ>0

No, one such sequence will do.

 

[Eclair_de_XII]

How about: $\{a_n: a_n=x-2^{-n}(x-a_{n-1}), \text{where} \space a_0\in (x-\epsilon,x) \cap S,\text{for} \space n\in ℕ\}$?

I read about a sequence like this for a proof for the Bolzano-Weierstrauss property in the book I'm using.

 

[haruspex]

$a_n=x-2^{-n}(x-a_{n-1})$?

Why should that be a member of S?

 

[Eclair_de_XII]

Let me try to explain this. I should wish my induction skills are up to par for this...

If $a_0\in (x-\epsilon,x)\cap S$, then $a_1=x-\frac{1}{2}(x-a_0)=\frac{1}{2}(x+a_0)>\frac{1}{2}(a_0+a_0)=a_0>x-\epsilon$. And since $x>a_0$, then $x=\frac{1}{2}(x+x)>\frac{1}{2}(x+a_0)=a_1>\frac{1}{2}(a_0+a_0)=a_0>x-\epsilon$. Thus, $a_1\in (a_0,x)\subset (x-\epsilon,x)\cap S$.

Now we suppose that $a_k \in (x-\epsilon,x)\cap S$ and $a_i>a_j \space \text{for} \space i>j$ for all $i,j \in ℕ$.
Then $x>a_k=x-\frac{1}{2^k}(x-a_{k-1})>a_0>x-\epsilon$.
We now consider $a_{k+1}=x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k})]>x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k-1})]>x-\frac{1}{2^k}(x-a_{k-1})>a_0>x-\epsilon$. Moreover, $x>x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k})]$, since $\frac{1}{2}[\frac{1}{2^k}(x-a_{k})]>0$.

Hence, every term of the sequence $\{a_n\}$ is bounded from above by $x=sup(S)$ and $a_0\in (x-\epsilon,x)\cap S$. The interval $(x-\epsilon,x)$ contains $\{a_n\}$ as a result.

 

[haruspex]

Thus, $...(a_0,x)\subset (x-\epsilon,x)\cap S$.

No. There is no reason at all why that interval should be a subset of S.
S could be countably infinite, no interval subsets.

 

[Eclair_de_XII]

So I'm thinking I should adjust my proof? I could just adjust my hypothesis to say that each term in $\{a_n\}$ is contained in $(x-\epsilon,x)$.

I've been thinking of just adjusting my induction proof to something like:

"We start by noting that since $x-\epsilon<sup(S)$, there exists $a_0 \in S$, such that $a_0>x-\epsilon$.

Suppose that for all $i>j\in ℕ\cup \{0\}$,$a_i>a_j$.

We proceed by induction. $a_1=x-\frac{1}{2}(x-a_0)=\frac{1}{2}(x+a_0)$. Noting that $x>a_0$, $x=\frac{1}{2}(x+x)>\frac{1}{2}(x+a_0)=a_1>\frac{1}{2}(a_0+a_0)=a_0>x-\epsilon$. Thus, for $n=1$, $a_1>a_0$; moreover, $x>a_1>x-\epsilon$.

We now suppose that $a_k=x-\frac{1}{2^k}(x-a_{k-1})>a_{k-1}$ for some $k\geq 2$. Then we consider $a_{k+1}=x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k})]=x+\frac{1}{2}[\frac{1}{2^k}(a_{k}-x)]> \\ x+\frac{1}{2}[\frac{1}{2^k}(a_{k-1}-x)]=x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k-1})]>x-\frac{1}{2^k}(x-a_{k-1})=a_k$ Thus, for all $i,j\in ℕ$, $a_i>a_j$.

Now, we observe that for any given $a_k$, $x=x-0>x-\frac{1}{2^k}(x-a_{k-1})=a_k$. Moreover, since $k\geq 2>0$, then $a_k>a_0>x-\epsilon$. So: $x>a_k>x-\epsilon$ for some $k\in ℕ\cup \{0\}$. The sequence $\{a_n\}$ is contained within $(x-\epsilon,x)$, as a result."

 

[haruspex]

So I'm thinking I should adjust my proof? I could just adjust my hypothesis to say that each term in $\{a_n\}$ is contained in $(x-\epsilon,x)$.

I've been thinking of just adjusting my induction proof to something like:

"We start by noting that since $x-\epsilon<sup(S)$, there exists $a_0 \in S$, such that $a_0>x-\epsilon$.

Suppose that for all $i>j\in ℕ\cup \{0\}$,$a_i>a_j$.

We proceed by induction. $a_1=x-\frac{1}{2}(x-a_0)=\frac{1}{2}(x+a_0)$. Noting that $x>a_0$, $x=\frac{1}{2}(x+x)>\frac{1}{2}(x+a_0)=a_1>\frac{1}{2}(a_0+a_0)=a_0>x-\epsilon$. Thus, for $n=1$, $a_1>a_0$; moreover, $x>a_1>x-\epsilon$.

We now suppose that $a_k=x-\frac{1}{2^k}(x-a_{k-1})>a_{k-1}$ for some $k\geq 2$. Then we consider $a_{k+1}=x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k})]=x+\frac{1}{2}[\frac{1}{2^k}(a_{k}-x)]> \\ x+\frac{1}{2}[\frac{1}{2^k}(a_{k-1}-x)]=x-\frac{1}{2}[\frac{1}{2^k}(x-a_{k-1})]>x-\frac{1}{2^k}(x-a_{k-1})=a_k$ Thus, for all $i,j\in ℕ$, $a_i>a_j$.

Now, we observe that for any given $a_k$, $x=x-0>x-\frac{1}{2^k}(x-a_{k-1})=a_k$. Moreover, since $k\geq 2>0$, then $a_k>a_0>x-\epsilon$. So: $x>a_k>x-\epsilon$ for some $k\in ℕ\cup \{0\}$. The sequence $\{a_n\}$ is contained within $(x-\epsilon,x)$, as a result."

This is getting much too complicated, and at a skim I would say it still has the same problem. You cannot construct members of S this way.

Try considering a sequence of ε_n converging to zero and base the a_n on those.

 

[Eclair_de_XII]

So do I establish a sequence of $\epsilon_n$ and $a_n$ based on that sequence?

 

[LCKurtz]

Try considering a sequence of ε_n converging to zero and base the a_n on those.

So do I establish a sequence of $\epsilon_n$ and $a_n$ based on that sequence?

What do you think?

 

[haruspex]

So do I establish a sequence of $\epsilon_n$ and $a_n$ based on that sequence?

Some confusion from the fonts, maybe. ε and $\epsilon$ are the same, just different fonts. Choose the ε_n (or $\epsilon_n$) sequence and base the a_n sequence on that.

 

[Eclair_de_XII]

The only sequence I can think of is $\{\epsilon_n:\epsilon_n=\frac{\epsilon}{n},n\geq N\in ℕ\}$, but that just brings back the problem from post #4...

 

[haruspex]

The only sequence I can think of is $\{\epsilon_n:\epsilon_n=\frac{\epsilon}{n},n\geq N\in ℕ\}$, but that just brings back the problem from post #4...

No, your problem in post #4 was that you assumed x-ε/n was a member of S. Your task now is to base a_n on $\epsilon_n=\frac{\epsilon}{n}$ such that a_n is in S.

 

[Eclair_de_XII]

Okay, I'm going to go with the sequence:

$\{a_n:a_n=a_0+\epsilon_n,\text {for} \space n \geq N \in ℕ, \text {where} \space a_0\in S\cap(x-\epsilon,x] \space \text {and} \space a_N\leq x\}$

 

[haruspex]

Okay, I'm going to go with the sequence:

$\{a_n:a_n=a_0+\epsilon_n,\text {for} \space n \geq N \in ℕ, \text {where} \space a_0\in S\cap(x-\epsilon,x] \space \text {and} \space a_N\leq x\}$

Same old problem. There is absolutely no reason to suppose that creates a_n as a member of S.

Let's go back to the start. You picked an ε and based on that you found a in such a way that it is guaranteed to be a member of S. Use the same method to base a_n on ε_n so that a_n is a member of S.

 

[Eclair_de_XII]

Assuming you mean my usage of the least-upper-bound property, then I should think that such a sequence would look something like this, perhaps?

Let $P=\{a_n: sup(S) = x\geq a_n > x-\epsilon_n,n \in ℕ\}$.

 

[haruspex]

Assuming you mean my usage of the least-upper-bound property, then I should think that such a sequence would look something like this, perhaps?

Let $P=\{a_n: sup(S) = x\geq a_n > x-\epsilon_n,n \in ℕ\}$.

Yes, but you cannot write it like that because that does not work as a definition of the a_n. A set definition in that form should tell the reader exactly how to find each member of the set.
You have to use the same approach as for a. Given ε_n > 0 ∃ a_n such that...

 

[Eclair_de_XII]

Let me retry: $\{a_n: \text {Given} \space \epsilon_n>0,\space ∃a_n \space \text {such that} \space sup(S)=x \geq a_n>x-\epsilon_n,\space \forall n\in ℕ\}$.

 

[haruspex]

Let me retry: $\{a_n: \text {Given} \space \epsilon_n>0,\space ∃a_n \space \text {such that} \space sup(S)=x \geq a_n>x-\epsilon_n,\space \forall n\in ℕ\}$.

Better, but still technically incorrect. If you use {x:P(x)} then it should be completely prescriptive. You cannot have choices going on inside. Drop the {} notation and just show that for each ε_n it is possible to choose an a_n with certain desirable properties.

 

[Eclair_de_XII]

"Because $x$ is the least upper bound of $S$ and for every $\epsilon_n=\frac{\epsilon}{n}>0$, where $n\in ℕ$, every $x-\epsilon_n$ is not an upper bound for $S$, since $x-\epsilon_n<sup(S)=x$ for all $\epsilon_n>0$. As a result, it is always possible to find an $a_n\in S$ such that $a_n>x-\epsilon_n$. Moreover, since $x$ is an upper bound for $S$, $x\geq a_n$ for all $a_n\in S$. Let $P=\{a_n:sup(S)=x\geq a_n>x-\epsilon_n, \space \forall n \in ℕ\}$ denote the collection of all $a_n$ with this property."

How is this? Oh, and I should probably show that the sequence converges to $x$, shouldn't I?

"Suppose that $\{a_n\}$ converges to $x$ for some $n\geq N\in ℕ$. Then $|x-a_n|<\epsilon'$ for all $\epsilon'>0$. Notice that $|x-(x-\epsilon_n)|>|x-a_n|$, since $a_n>x-\epsilon_n$ and $-a_n<-(x-\epsilon_n)$. We choose $\epsilon'=\frac{\epsilon}{N}$ so that $\epsilon'=\frac{\epsilon}{N}\geq \frac{\epsilon}{n}=|x-(x-\epsilon_n)|=|\epsilon_n|=|\frac{\epsilon}{n}|>|x-a_n|$."

 

[Eclair_de_XII]

Well, I think I have enough information to do the rest of my homework now. Thanks everyone, for being patient with me.

 

[haruspex]

Let $P=\{a_n:sup(S)=x\geq a_n>x-\epsilon_n, \space \forall n \in ℕ\}$ denote the collection of all $a_n$ with this property.

You were doing well until there. That set turns out be just {x}, which need not be a member of S. But you do not need to define a set. You need to define a sequence, and you already did that in the preceding sentences.

Suppose that $\{a_n\}$ converges to $x$ for some $n\geq N\in ℕ$.

You cannot go around supposing that. You have to show that it converges (which is quite easy).