Proof: A point is a limit point of S is a limt of a sequence

[zigzagdoom]

Hi guys,

I attempted to prove this theorem, but just wanted to see if it a valid proof.

Thanks!

1. Homework Statement

Prove that x is an accumulation point of a set S iff there exists a sequence ( s _n ) of points in S \ {x} that converges to x

Homework Equations

N * ( x; ε ) is the x - deleted ε - neighbourhood of x

The Attempt at a Solution

Suppose that x is an accumulation point of a set S. Then N * ( x; ε ) ⋀ S ≠ / Ø

Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø .

Since y ∈ N * ( x; ε ) , 0 < | y - x | < ε , which means | y - x | < ε for any y. Now consider any sequence of y ∈ N * ( x; ε ) and call this ( s _n ).

For every n > N, where N can be any number of the index, |( s _n ) - x|< ε which means ( s _n ) converges to x.

Conversely, suppose that there exists a sequence ( s _n ) of points in S \ {x} that converges to x .

Then for ∀ ε >0, ∃ N such that ∀ n>N, it is the case that |( s _n )-x|< ε . But this is the ε - neighbourhood of x, so that ( s _n ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point

 

[HallsofIvy]

No, this is not valid because the "$N(x,\epsilon)$" you are talking about is for a specific $\epsilon$. There is no reason to think that the $s_n$ get close to x. Instead, look at all the $N(x, \epsilon)$ with $\epsilon= 1/n$, n= 1, 2, 3, ...

 

[fresh_42]

How is accumulation point defined here? What is S? Which metric do you use, if at all? In which space is the question located? ℝ? How can you conclude that $N^*(x;ε) ≠ ∅$?

 

[HallsofIvy]

I believe zigzagdoom said, or at least implied, the definition of "accumulation" point: "p is an accumulation point of set S if and only if every neighborhood of x contains at least one point, other than x itself, of S- equivalently, that every "deleted" neighbor hood of x (a neighborhood of x with x "deleted" (removed from the set). A "neighborhood" of x in a metric space is $N(x, \epsilon)= \{ p| d(x, p)< \epsilon\}$. A "deleted neighborhood" is $N*(x, \epsilon)= \{p| 0< d(p, x)< \epsilon\}$. This is true for all metric spaces so it is not necessary nor desirable to specify a metric or the space.[/B]

 

[fresh_42]

This is true for all metric spaces so it is not necessary nor desirable to specify a metric or the space.

Yep. But Hausdorff would do. Without being metric. I remember a textbook in physics where a complete space has been defined by: "... when for every x there is a sequence converging to x."
However, whether at school or in science: you can never be too young to get used to state problems in a proper way. That often leads to a much clearer understanding and may help to reveal the solution by itself. Plus it avoids almost always mistakes. (personal opinion)

 

[zigzagdoom]

No, this is not valid because the "$N(x,\epsilon)$" you are talking about is for a specific $\epsilon$. There is no reason to think that the $s_n$ get close to x. Instead, look at all the $N(x, \epsilon)$ with $\epsilon= 1/n$, n= 1, 2, 3, ...

Thanks I will rework and see if I can incorporate this

 

[zigzagdoom]

How is accumulation point defined here? What is S? Which metric do you use, if at all? In which space is the question located? ℝ? How can you conclude that $N^*(x;ε) ≠ ∅$?

Thanks,

In response to your questions: x is an accumulation point if N(x;ε) ∧ S ≠ ∅. S is not defined, but I have taken it to mean some metric space (the metric has not been specified). The space of the question has not been specified. I have assumed x is an accumulation point, so that N(x;ε) ∧ S ≠ ∅(by definition).

 

[mfb]

In response to your questions: x is an accumulation point if N(x;ε) ∧ S ≠ ∅

For ever real ε>0. This is an important part of the definition.

 

[fresh_42]

Thanks,

In response to your questions: x is an accumulation point if N(x;ε) ∧ S ≠ ∅. S is not defined, but I have taken it to mean some metric space (the metric has not been specified). The space of the question has not been specified. I have assumed x is an accumulation point, so that N(x;ε) ∧ S ≠ ∅(by definition).

Sorry I read ∧ as AND. And one reason I was asking is, I looked it up to remember its exact definition and to avoid mistakes. I found your problem above as such in the case S is any sequence. This left me alone with the question what your definition is. If I may try to understand it better: x is an accumulation point of S iff for every ε > 0 there can be found an open neighborhood N(x;ε) of x such that (S ∩ N(x;ε)) \ {x} ≠ ∅.
One has to be precise here because the gap between the definition and what has to be shown is rather close. It's almost the same.

 

[zigzagdoom]

.

Hi guys,

I attempted to prove this theorem, but just wanted to see if it a valid proof.

Thanks!

1. Homework Statement

Prove that x is an accumulation point of a set S iff there exists a sequence ( s _n ) of points in S \ {x} that converges to x

Homework Equations

N * ( x; ε ) is the x - deleted ε - neighbourhood of x if for every ε>0, N * ( x;ε) ⋀ S ≠ Ø.

The Attempt at a Solution

(version 2)[/B]

Suppose that x is an accumulation point of a set S. Then N * ( x; ε ) ⋀ S ≠ Ø

Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø .

Since y ∈ N * ( x; ε ) , 0 < | y - x | < ε , which means | y - x | < ε for any y. Now consider any sequence of y ∈ N * ( x; ε ) and call this ( s _n ).

For every n > N, where N can be any number of the index, |( s _n ) - x|< ε which means ( s _n ) converges to x.

Conversely, suppose that there exists a sequence ( s _n ) of points in S \ {x} that converges to x .

Then for ∀ ε >0, ∃ N such that ∀ n>N, it is the case that |( s _n )-x|< ε . But this is the ε - neighbourhood of x, so that ( s _n ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point

Sorry I read ∧ as AND. And one reason I was asking is, I looked it up to remember its exact definition and to avoid mistakes. I found your problem above as such in the case S is any sequence. This left me alone with the question what your definition is. If I may try to understand it better: x is an accumulation point of S iff for every ε > 0 there can be found an open neighborhood N(x;ε) of x such that S ∩ N(x;ε) ≠ ∅.
One has to be precise here because the gap between the definition and what has to be shown is rather close. It's almost the same.

Thanks again.

The precise definition that I have in the book is:

"x is an accumulation point of S if every deleted neighbourhood of x contains a point of S. That is, for every ε > 0, N * ( x;ε ) ≠ ∅."

N * ( x;ε ) ≠ ∅ is the deleted neighbourhood of x.

 

[fresh_42]

I've corrected my definition:
$x$ is an accumulation point of $S$ if and only if for every $ε > 0$ there can be found an open neighborhood $N(x;ε)$ of $x$ such that $S ∩ N^*(x;ε) ≠ ∅$
Ok, now you can apply Hallsoflvy's advice in post #2 and create your convergent sequence.

 

[zigzagdoom]

No, this is not valid because the "$N(x,\epsilon)$" you are talking about is for a specific $\epsilon$. There is no reason to think that the $s_n$ get close to x. Instead, look at all the $N(x, \epsilon)$ with $\epsilon= 1/n$, n= 1, 2, 3, ...

I've corrected my definition:
$x$ is an accumulation point of $S$ if and only if for every $ε > 0$ there can be found an open neighborhood $N(x;ε)$ of $x$ such that $S ∩ N^*(x;ε) ≠ ∅$
Ok, now you can apply Hallsoflvy's advice in post #2 and create your convergent sequence.

Second attempt at proof

Suppose that x is an accumulation point of a set S. Then for every ε > 0, it is the case that N * ( x; ε ) ⋀ S ≠ Ø.

Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø.

Now let a sequence of a number of y be denoted by ( s _n ). Construct the sequence by letting ϵ = 1/n, where n= 1, 2, 3, ... and picking each point within the sequence so that N * ( x;ϵ ) contains each successive member y of (s_n).

Then we have that for every ε we select, ∃N such that for all n > N, |(s_n) - x| < ε. This means ( s n ) converges to x.

Conversely, suppose that there exists a sequence ( s _n ) of points in S \ {x} that converges to x .

Then for ∀ ε >0, ∃ N such that for all n>N, it is the case that | ( s _n ) - x | < ε . But this is the ε - neighbourhood of x, so that ( s_n ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point.

 

[fresh_42]

Second attempt at proof

Suppose that x is an accumulation point of a set S. Then for every ε > 0, it is the case that N * ( x; ε ) ⋀ S ≠ Ø.

correct

Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø.

You don't consider any $y$, you choose such an $y_ε$ out of $N^* ( x; ε ) ∩ S$ which you know exists. Out of these $y_ε$ you construct a convergent sequence using Hallsofly's narrowing down $ε$.

Construct the sequence by letting ϵ = 1/n, where n= 1, 2, 3, ... and picking each point

$s_n$ in N^*(x;ε) ∩ S.

Then we have that for every ε

> 0 if

we select

N = next number above 1/ε

for all n > N: |(s_n) - x| <

1/n < 1/N

< ε. This means ( s_n ) converges to x.

Conversely, suppose that there exists a sequence ( s _n ) of points in S \ {x} that converges to x .

Then for ∀ ε >0, ∃ N such that for all n>N, it is the case that | s _n - x | < ε

But this is the ε - neighbourhood of x, so that ( s_n ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point.

correct. A little bit bumpy, but ok. Your $s_n$ are already in N^*(x;ε) ∩ S making it non empty and that's all you wanted to show. Forget about the y.

 

[zigzagdoom]

correct
You don't consider any $y$, you choose such an $y_ε$ out of $N * ( x; ε ) ∩ S$ which you know exists. Out of these $y_ε$ you construct a convergent sequence using Hallsofly's narrowing down $ε$.$s_n$ in N^*(x;ε) ∩ S.

> 0 if

N = next number above 1/ε

1/n < 1/Ncorrect. A little bit bumpy, but ok. Your $s_n$ are already in N^*(x;ε) ∩ S making it non empty and that's all you wanted to show. Forget about the y.

Thanks a lot for this - guidance has been very useful in helping me think about the problem.

 

[fresh_42]

Never mind. But you see the two definitions aren't far apart. That's why you have to be precise. One can easily mess things up.

 

[fresh_42]

A "neighborhood" of x in a metric space is N(x,ϵ)={p|d(x,p)<ϵ}N(x, \epsilon)= \{ p| d(x, p)< \epsilon\}. A "deleted neighborhood" is N∗(x,ϵ)={p|0<d(p,x)<ϵ}N*(x, \epsilon)= \{p| 0< d(p, x)< \epsilon\}. This is true for all metric spaces so it is not necessary nor desirable to specify a metric or the space.

Just out of curiosity. In the discrete metric $d(x,y) = 0$ for $x = y$ and $d(x,y) = 1$ for $x ≠ y$ there won't be any accumulation points, right? Ok, that does not contradict what you said but it would change the proof dramatically.

 

[mfb]

There are no accumulation points in that metric, right. There are also no series in S\{x} converging to x, for any x in S.

 

[fresh_42]

There are no accumulation points in that metric, right. There are also no series in S\{x} converging to x, for any x in S.

... which makes the proof rather short because all elements of the empty set have blue eyes.

 

[mfb]

This special case would be easier to show than the general case, sure.

@zigzagdoom: Don't forget the other direction.

 

[fresh_42]

This special case would be easier to show than the general case, sure.

@zigzagdoom: Don't forget the other direction.

But he had both directions. Did I overlook something?

 

[mfb]

Ah, I missed that part in the first post, sorry.