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).