0 is a cluster point of E^c

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G-X

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We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}

Let [Math]k = 1[/Math] or [Math]2[/Math] then by the definition above [Math]E^{c} = [/Math]{[Math]m \in R^{k}[/Math] : [Math]\forall m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]}

From here, we can define a cluster point and, apply it to [Math]E^{c}[/Math]. A point x is a cluster point of a set [Math]E[/Math] in [Math]R^{k}[/Math] where [Math]k = 1[/Math] or [Math]2[/Math], if for each [Math]r > 0[/Math], [Math]B_{r}(x)[/Math] contains infinitely many points of E, that is, [Math]B_{r}(x) \cap E[/Math] is an infinite set.

So we must show that for each [Math]r > 0[/Math], [Math]Br(0) \cap E^{c}[/Math] is infinite to prove that 0 is a cluster point of [Math]E^{c}[/Math].

Case 1: [Math]R^{1}[/Math] [Math]< 0[/Math]:

Case 2: [Math]R^{1}[/Math] [Math]> 0[/Math]:

Case 3: [Math]R^{1} = -1[/Math]:

Case 4: [Math]R^{1}[/Math] [Math]\in[/Math] (-1, 0):


Here I am a little confused because [Math]R^2[/Math] is no longer a number in [Math]R^1[/Math], so do the cases change to some like the below? Do you put it in terms of d(R^1, R^1)?

Case 1: [Math]R^{2}[/Math] [Math]< (0, 0)[/Math]:

Case 2: [Math]R^{2}[/Math] [Math]> (0, 0)[/Math]:

Case 3: [Math]R^{2} = (-1, 0)[/Math]:

Case 4: [Math]R^{2}[/Math] [Math]\in[/Math] {x: -1< x < 0, y: -1< y < 0} - Is this correct for a range similar to Case 4 above
 
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Figured out the answer to my own question: If this problem was to exist in [Math]R^{2}[/Math] then the original set would have to be rewritten from:
[Math]E = \{1 / n^{2} : n \in Z+\}[/Math]to
[Math]E = \{(1 / n^{2}, 0) : n \in Z+\} \subset R^{2}[/Math]thus
[Math]E = \{1 / n^{2} : n \in Z+\} \subset R^{1}[/Math] for this problem to be solvable.

We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}

For this problem to be solvable, [Math]E^{c} \subset R^{1}[/Math], so by the definition above:

[Math]E^{c} = [/Math]{[Math]m \in R^{1}[/Math] : [Math]\forall m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]} - this must be modified

From here, we can define a cluster point and, apply it to [Math]E^{c}[/Math]. A point x is a cluster point of a set [Math]E[/Math] in [Math]R^{k}[/Math] where [Math]k = 1[/Math] or [Math]2[/Math], if for each [Math]r > 0[/Math], [Math]B_{r}(x)[/Math] contains infinitely many points of E, that is, [Math]B_{r}(x) \cap E[/Math] is an infinite set.

So we must show that for each [Math]r > 0[/Math], [Math]Br(0) \cap E^{c}[/Math] is infinite to prove that 0 is a cluster point of [Math]E^{c}[/Math].

Case 1: [Math]R^{1}[/Math] [Math]< 0[/Math]:

Case 2: [Math]R^{1}[/Math] [Math]> 0[/Math]:

Case 3: [Math]R^{1} = -1[/Math]:

Case 4: [Math]R^{1}[/Math] [Math]\in[/Math] (-1, 0):
 
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G-X said:
We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}

Let [Math]k = 1[/Math] or [Math]2[/Math] then by the definition above [Math]E^{c} = [/Math]{[Math]m \in R^{k}[/Math] : [Math]\forall m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]}

From here, we can define a cluster point and, apply it to [Math]E^{c}[/Math]. A point x is a cluster point of a set [Math]E[/Math] in [Math]R^{k}[/Math] where [Math]k = 1[/Math] or [Math]2[/Math], if for each [Math]r > 0[/Math], [Math]B_{r}(x)[/Math] contains infinitely many points of E, that is, [Math]B_{r}(x) \cap E[/Math] is an infinite set.

So we must show that for each [Math]r > 0[/Math], [Math]Br(0) \cap E^{c}[/Math] is infinite to prove that 0 is a cluster point of [Math]E^{c}[/Math].

Case 1: [Math]R^{1}[/Math] [Math]< 0[/Math]:

Case 2: [Math]R^{1}[/Math] [Math]> 0[/Math]:

Case 3: [Math]R^{1} = -1[/Math]:

Case 4: [Math]R^{1}[/Math] [Math]\in[/Math] (-1, 0):


Here I am a little confused because [Math]R^2[/Math] is no longer a number in [Math]R^1[/Math], so do the cases change to some like the below? Do you put it in terms of d(R^1, R^1)?

Case 1: [Math]R^{2}[/Math] [Math]< (0, 0)[/Math]:

Case 2: [Math]R^{2}[/Math] [Math]> (0, 0)[/Math]:

Case 3: [Math]R^{2} = (-1, 0)[/Math]:

Case 4: [Math]R^{2}[/Math] [Math]\in[/Math] {x: -1< x < 0, y: -1< y < 0} - Is this correct for a range similar to Case 4 above
Click to expand...
I think that the only thing that changes is the metric together with your definition of Br(x). Cluster point is a topological definition that has almost nothing to do with the dimension of your space.
 
G-X said:
We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}

Let [Math]k = 1[/Math] or [Math]2[/Math] then by the definition above [Math]E^{c} = [/Math]{[Math]m \in R^{k}[/Math] : [Math]\forall m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]}

From here, we can define a cluster point and, apply it to [Math]E^{c}[/Math]. A point x is a cluster point of a set [Math]E[/Math] in [Math]R^{k}[/Math] where [Math]k = 1[/Math] or [Math]2[/Math], if for each [Math]r > 0[/Math], [Math]B_{r}(x)[/Math] contains infinitely many points of E, that is, [Math]B_{r}(x) \cap E[/Math] is an infinite set.

So we must show that for each [Math]r > 0[/Math], [Math]Br(0) \cap E^{c}[/Math] is infinite to prove that 0 is a cluster point of [Math]E^{c}[/Math].

Case 1: [Math]R^{1}[/Math] [Math]< 0[/Math]:

Case 2: [Math]R^{1}[/Math] [Math]> 0[/Math]:

Case 3: [Math]R^{1} = -1[/Math]:

Case 4: [Math]R^{1}[/Math] [Math]\in[/Math] (-1, 0):


Here I am a little confused because [Math]R^2[/Math] is no longer a number in [Math]R^1[/Math], so do the cases change to some like the below? Do you put it in terms of d(R^1, R^1)?

Case 1: [Math]R^{2}[/Math] [Math]< (0, 0)[/Math]:

Case 2: [Math]R^{2}[/Math] [Math]> (0, 0)[/Math]:

Case 3: [Math]R^{2} = (-1, 0)[/Math]:

Case 4: [Math]R^{2}[/Math] [Math]\in[/Math] {x: -1< x < 0, y: -1< y < 0} - Is this correct for a range similar to Case 4 above
Click to expand...
You clearly have not stated the problem as given to you, because you didn't define E before writing its complement, and because [MATH]m \neq \frac{1}{m^{2}}[/MATH] can't be right. That would just mean m = 1 is not a positive integer. We need to see the actual entire problem.

Also, statements like [MATH]R^1<0[/MATH] are nonsense, as [MATH]R^1[/MATH] is an (infinite) set, not a number.
 
Dr. Peterson - Thank you for pointing that out. That makes sense.

Also to answer an incomplete understanding: Those above cases attempt to prove that there are no other cluster points. We only need a single case and that is for z = 0.

It seems there would be a way to show that between any r for [Math]r1 < r2 < r3[/Math] there will be only finite points for a function like [Math]1 /n^2[/Math] since [Math]n \in Z+[/Math]. I believe that if n had been in [Math]R^1[/Math] then the intersection would be infinite.

If we can prove that there exists a [Math]r'[/Math] where [Math]Br'(0) ⋂ E[/Math] is infinite then it implies that if [Math]r' < r[/Math] that means r contains all the points of r' and is thus infinite? Yes.

So we must show that the points contained by r1 is infinite to solve the problem. On the other hand if I want to prove that there are finite points - do I continue by induction of some form? That r3 contains finite points, r2 contains finite points then by maybe strong induction r1 is finite? Thus the intersection is finite and 0 is not a cluster point.
 
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G-X said:
We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}
Let [Math]k = 1[/Math] or [Math]2[/Math] then by the definition above [Math]E^{c} = [/Math]{[Math]m \in R^{k}[/Math] : [Math]\forall m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]}
Click to expand...
I have stayed away from this thread because I did not know where to start to correct your confusion.
Here are three books:
1) SET THEORY by Feilix Hausdorff is the basic reference for all interested in this material.
2) General Topology by John L. Kelly is a standard textbook in topology courses in North American Colleges.
3) Elementary Theory of Metric Spaces by Robert B. Reisel is a text that I have used with in-service teachers. In is designed as a Moore type course in constructing mathematical proofs.
Now here are the corrections that I think you need to consider.
A sequence is a function from the set of natural numbers,\(\mathbb{N}\), into some metric space.(That is generally the case)
We say that a sequence \(a_n\) converges to a number \(L\) proved that the ball \(\mathcal{B}_{\delta}(L)\) contains almost all terms of \(a_n\).
Why did I empathize the term almost all ? It means "all but a finite collection of the terms of \(a_n\). So every ball that contains \(L\) contains infinity many terms of \(a_n\). At this point we need to remind all that a sequence is a countable set.

Now we turn to a general idea of limit point. Suppose that \(\mathcal{J}\) is a subset of a metric space \(\mathcal{X}\).
Again realize that there is a metric \(d\) and basic open sets are balls \(\delta>0,\:\:\mathcal{B}_{\delta}(P)=\{T: d(P,T)<\delta\}\)
Technically, the statement that \(L\) is a limit point of the set \( \mathcal{J}\) means for any ball \(\mathcal{B}_{\delta}(L)\cap\left(\mathcal{J}\setminus\{L\}\right)\ne\emptyset\).
The last bit is hard to follow, so lets simplify it. The statement that \(L\) is a limit point of a set \(\mathcal{J}\) means the any open set that contains the limit point \(L\) also contains a point of the set \(\mathcal{J}\) different from \(L\).
From that last bit if follows that only infinite sets can have limits points.
 
G-X said:
We can define the complement, [Math]E^{c}[/Math], of a subset E of [Math]R^{k}[/Math], k = 1 or 2, is the set of all elements of [Math]R^{k}[/Math] which are not in E, that is, [Math]E^{c} = [/Math]{[Math]x \in R^{k}[/Math]: [Math]x \notin E[/Math]}

Let [Math]k = 1[/Math] or [Math]2[/Math] then by the definition above [Math]E^{c} = [/Math]{[Math]m \in R^{k}[/Math] : [Math]\for all m[/Math] [Math]\in Z+, m \neq \frac{1}{m^{2}} [/Math]}
Click to expand...
This is true if \(\displaystyle E= \{m\in Z^+| m= \frac{1}{m^2}\}\)- but you haven't defined E!

From here, we can define a cluster point and, apply it to [Math]E^{c}[/Math]. A point x is a cluster point of a set [Math]E[/Math] in [Math]R^{k}[/Math] where [Math]k = 1[/Math] or [Math]2[/Math], if for each [Math]r > 0[/Math], [Math]B_{r}(x)[/Math] contains infinitely many points of E, that is, [Math]B_{r}(x) \cap E[/Math] is an infinite set.

So we must show that for each [Math]r > 0[/Math], [Math]Br(0) \cap E^{c}[/Math] is infinite to prove that 0 is a cluster point of [Math]E^{c}[/Math].

Case 1: [Math]R^{1}[/Math] [Math]< 0[/Math]:

Case 2: [Math]R^{1}[/Math] [Math]> 0[/Math]:

Case 3: [Math]R^{1} = -1[/Math]:

Case 4: [Math]R^{1}[/Math] [Math]\in[/Math] (-1, 0):
Click to expand...
?? "[math]R^1[/math] is the set of all real numbers, not a number so these make no sense!


Here I am a little confused because [Math]R^2[/Math] is no longer a number in [Math]R^1[/Math], so do the cases change to some like the below? Do you put it in terms of d(R^1, R^1)?

Case 1: [Math]R^{2}[/Math] [Math]< (0, 0)[/Math]:

Case 2: [Math]R^{2}[/Math] [Math]> (0, 0)[/Math]:

Case 3: [Math]R^{2} = (-1, 0)[/Math]:

Case 4: [Math]R^{2}[/Math] [Math]\in[/Math] {x: -1< x < 0, y: -1< y < 0} - Is this correct for a range similar to Case 4 above
Click to expand...
Again, none of these make sense because [math]R^2[/math], like [math]R^1[/math], is a set, not a number.
 
I was able to finish the proof - I believe. The solution I arrived at was quite long. You should break up the cases depending upon where Br(0) exists, R, Q, etc.

Do not take my word for it though.

I will take a look into those supplementary materials. Thank you for sharing.
 
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