# Convergence of sequence in metric space proof

• Incand
In summary: I think I see what you mean here. Accodring to the definition i needed to show for every neighbourhood but I only showed for open balls. However since i can always find an open ball that is contained in that neighbourhood the proof is fine anyway since ##p_n \in B(p,\epsilon) \subseteq N(p)## for some ##\epsilon##. Is that right? I think you have it.
Incand

## Homework Statement

Let ##E \subseteq M##, where ##M## is a metric space.
Show that
##p\in \overline E = E\cup E' \Longleftrightarrow## there exists a sequence ##(p_n)## in ##E## that converges to ##p##.

##E'## is the set of limit points to ##E## and hence ##\overline E## is the closure of ##E##.

## Homework Equations

Definition:[/B]
A point ##p## is a limit point to ##E## if ##\forall \epsilon > 0## there is a point ##q\in B(p,\epsilon)## so that ##q\in E## and ##q \neq p##.

Definition:
##(p_n)## converges if ##\forall \epsilon > 0## exists ##N## so that
##d(p,p_n)< \epsilon## for all ##n\ge N##.

## The Attempt at a Solution

I think I managed to show the above but I would like someone to go over my proof and check if it's correct:

Starting with ##\Longleftarrow## we have ##\forall \epsilon > 0##
##d(p_n,p)<\epsilon## ##\forall n \ge N## for some ##N## where ##p_n \in E##. But this means that ##p## is a limit point of ##E##. i.e. ##p \in E' \subseteq \bar E##.For ##\Longrightarrow## we have ##p \in \overline E##.
If ##p \in E## the sequence of ##p_n = p## does it.

If ##p \in E'## we select
##p_n \in B(p,1/n)## for ##n \in Z^+## so that ##p_n \in E##.
But then ##d(p_n,p)<1/n = \epsilon## for all ##n\ge N_\epsilon = 1/\epsilon##. i.e. ##p_n \to p##.

One question I have is if the detour around ##1/n## is neccesary or if I simply could've picked ##p_n \in B(p,\epsilon)##.

The second part of the proof uses the axiom of choice (AC). It is preferred to not use except when necessary, in order to gain an understanding of which results rely on AC and which do not. The place where it uses it is here:
If ##p \in E'## we select
##p_n \in B(p,1/n)## for ##n \in Z^+## so that ##p_n \in E##.
I'm pretty sure AC is not necessary for this proof, but whether it is depends on your definition of limit point (see below).

Also, I am having trouble parsing the first part into a meaningful mathematical statement. It seems to be missing a verb or some punctuation but in any case it needs some adjustment to be clearly comprehensible:
Starting with ##\Longleftarrow## we have ##\forall \epsilon > 0##
##d(p_n,p)<\epsilon## ##\forall n \ge N## for some ##N## where ##p_n \in E##.
I suspect you are not far from having an acceptable proof, but to be sure we need to know exactly what your definition of 'limit point' is, as subtly different definitions are available, and those differences can make some hard bits easy and other easy bits hard.

Incand
The exact statement of limit point in our book is:
Let ##X## be a metric space, All points and sets mentioned are understood to be elements and subsets of ##X##.
A point ##p## is a limit point of the set ##E## if every neighborhood of ##p## contains a point ##q \ne p## such that ##q \in E##.

The book never defines neighborhood, however we take it to be an open ball around the point i.e. ##B(p,\epsilon) = \{ x\in X, d(p,x)< \epsilon \}##.

I'm not sure how to get around using AC. I think a lot of proofs in our course actually uses that without mentionen it by name if that is what allows us to select elements from infinite sets. Or perhaps there was something particular I did i my proof beyond that?

For the first part I just wrote down the definition of convergence of a sequence which seemed to make the implication trivial.
The sequence being in ##E## means that ##p_n \in E##. That the sequence converges to ##p## means that
##\forall \epsilon > 0 \; \exists N## so that ##d(p,p_n)< \epsilon## ##\forall n \ge N##

But as long as one of those ##p_n \ne p ## (satisfying the above) this means exactly that ##p_n \in B(p,\epsilon)## i.e. ##p## is a limit point since this ##p_n \in E##.
The only possibility for not finding such a ##p_n \ne p## is if for all ##n## ##p_n = p## but then ##p \in E## so we're fine either way.

The argument in the first part is just missing an argument about 'every neighbourhood', which is understandable if you've never had 'neighbourhood' defined. In a metric space (or, more generally, in a topological space) a neighbourhood of a point is an open set containing the point*. You should have a theorem that tells you that if ##p\in U## for ##U## an open set, there must exist some ##\epsilon>0## such that the ball ##B_\epsilon(p)\subseteq U##. That should equip you to complete the proof of the first part.

Regarding the other direction, I can confirm that the axiom of choice is needed. See this stack exchange response for more detail. The approach you've taken there, using balls of radius ##1/n## is fine. I don't think there is a shorter way to do it.

* Sometimes a different definition of 'neighbourhood' as a set - not necessarily open - containing an open set containing the point, is used. But that difference rarely affects any proofs, and doesn't affect this one since, in the definition of 'limit point', if the requirement upon 'every neighbourhood of the point' holds when neighbourhood is interpreted as a set containing an open set containing the point, then it will also hold for every open set containing the point.

Incand
andrewkirk said:
The argument in the first part is just missing an argument about 'every neighbourhood', which is understandable if you've never had 'neighbourhood' defined. In a metric space (or, more generally, in a topological space) a neighbourhood of a point is an open set containing the point*. You should have a theorem that tells you that if ##p\in U## for ##U## an open set, there must exist some ##\epsilon>0## such that the ball ##B_\epsilon(p)\subseteq U##. That should equip you to complete the proof of the first part.
I think I see what you mean here. According to the definition i needed to show for every neighbourhood but I only showed for open balls. However since i can always find an open ball that is contained in that neighbourhood the proof is fine anyway since ##p_n \in B(p,\epsilon) \subseteq N(p)## for some ##\epsilon>0##.

I think part of the confusion were that our book speaks about neighbourhoods while in the courses in calculus/analysis I've taken we always used open balls.

andrewkirk said:
Regarding the other direction, I can confirm that the axiom of choice is needed. See this stack exchange response for more detail. The approach you've taken there, using balls of radius ##1/n## is fine. I don't think there is a shorter way to do it.

That's good to know! Thanks for all the help!

## 1. What is the definition of convergence in a metric space?

In a metric space, a sequence of points converges to a limit if the distance between the points in the sequence and the limit approaches zero as the sequence progresses.

## 2. How is convergence of a sequence in a metric space proven?

To prove convergence of a sequence in a metric space, one must show that for any given distance epsilon, there exists an index N such that all points in the sequence after N are within epsilon distance from the limit.

## 3. Can a sequence converge to multiple limits in a metric space?

No, in a metric space, a sequence can only converge to a single limit.

## 4. What is the role of the metric in the proof of convergence of a sequence in a metric space?

The metric defines the distance between points in a metric space and is used to determine the convergence of a sequence. The proof relies on the properties of the metric, such as the triangle inequality, to show that the sequence approaches the limit within a given distance.

## 5. Are all convergent sequences in a metric space Cauchy sequences?

Yes, in a metric space, all convergent sequences are also Cauchy sequences. This means that as the sequence progresses, the points become arbitrarily close to each other, leading to convergence to a limit.

• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
700
• Calculus and Beyond Homework Help
Replies
1
Views
729
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
391
• Calculus and Beyond Homework Help
Replies
2
Views
439
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Differential Equations
Replies
1
Views
1K