Normed Linear Space - Query

In summary, the conversation discusses the concept of convergence sequences being unique in a normed linear space. The participants discuss how this concept is true in every normed space and provide hints on how to prove it using the definition of convergence. The conversation also touches on the importance of understanding the definition of a norm and the concept of non-negative numbers in completing the proof. Additionally, the possibility of this concept being valid in certain other spaces is mentioned.
  • #1
bugatti79
794
1
Folks,

Can anyone give an example of where convergence sequences can be unique in a normed linear space?

THanks
 
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  • #2
What do you mean with "convergence sequences are unique"?? I don't really understand what you're asking.
 
  • #3
micromass said:
What do you mean with "convergence sequences are unique"?? I don't really understand what you're asking.

if x_n is a sequence and converges to x and y then x=y. THis is what I am looking for.
 
  • #4
bugatti79 said:
if x_n is a sequence and converges to x and y then x=y. THis is what I am looking for.

Yes, that's true in every normed space. What are you looking for?? A proof of that?
 
  • #5
micromass said:
Yes, that's true in every normed space. What are you looking for?? A proof of that?

yes if it is standard textbook stuff. Its not in my book. Thanks !
 
  • #6
OK, I'll give you a hint on how to prove it:

[tex]\|x-y\|\leq \|x-x_n\|+\|x_n-y\|[/tex]

Use the definition of convergence of a sequence now.
 
  • #7
micromass said:
OK, I'll give you a hint on how to prove it:

[tex]\|x-y\|\leq \|x-x_n\|+\|x_n-y\|[/tex]

Use the definition of convergence of a sequence now.

How this you arrive with the right hand side of the expression? Would the line before that have been

[tex]\|x-y\|\leq \|(x-x_n)+(x_n-y)\|[/tex]
 
  • #8
Yes, but with the = sign, since you're adding and subtracting the same number inside the norm.
 
  • #9
micromass said:
OK, I'll give you a hint on how to prove it:

[tex]\|x-y\|\leq \|x-x_n\|+\|x_n-y\|[/tex]

Use the definition of convergence of a sequence now.

ok,

[tex]\|x-y\|\leq \|(x-x_n)+(x_n-y)\|[/tex]

[tex]\|x-y\|= \|(x-x_n)+(x_n-y)\| \le \|x-x_n\|+\|x_n-y\| \le \|-(x_n-x)\|+\|x_n-y\| [/tex]

I have rewritten the above because we need the expressions in the form ||xn-x||=0 as n tends to infinity

now,
since xn=x for the limit n to infinity and and given ε>0 then ε/2>0

therefore so there exist [tex]n_1 ∈ N[/tex] such that

[tex]\|-(x_n-x)\|<ε/2[/tex] and [tex]n_2 ∈ N[/tex]

[tex]\|(x_n-y)\|<ε/2[/tex]

How I doing so far?
 
  • #10
bugatti79 said:
ok,

[tex]\|x-y\|\leq \|(x-x_n)+(x_n-y)\|[/tex]

[tex]\|x-y\|= \|(x-x_n)+(x_n-y)\| \le \|x-x_n\|+\|x_n-y\| \le \|-(x_n-x)\|+\|x_n-y\| [/tex]

I have rewritten the above because we need the expressions in the form ||xn-x||=0 as n tends to infinity

now,
since xn=x for the limit n to infinity and and given ε>0 then ε/2>0

therefore so there exist [tex]n_1 ∈ N[/tex] such that

[tex]\|-(x_n-x)\|<ε/2[/tex] and [tex]n_2 ∈ N[/tex]

[tex]\|(x_n-y)\|<ε/2[/tex]

How I doing so far?

[tex]n_1 ∈ N[/tex]

The third line back up...I think I can take out the minus sign giving


[tex]-\|(x_n-x)\|<ε/2[/tex] and [tex]n_2 ∈ N[/tex]

[tex]\|(x_n-y)\|<ε/2[/tex]...Any suggestions on what to do next?
 
  • #11
bugatti79 said:
since xn=x for the limit n to infinity and and given ε>0 then ε/2>0

therefore so there exist [tex]n_1 ∈ N[/tex] such that

[tex]\|-(x_n-x)\|<ε/2[/tex] and [tex]n_2 ∈ N[/tex]

[tex]\|(x_n-y)\|<ε/2[/tex]

How I doing so far?
Not bad, but there are some inaccuracies. You should start by saying something like "let ε>0 be arbitrary". Then "Since [itex]\lim x_n=x[/itex], there's an [itex]n_1\in\mathbb N[/itex] such that..." You also left out something really essential in that step. You didn't say that the first inequality holds for all [itex]n\in\mathbb N[/itex] such that [itex]n\geq n_1[/itex].

bugatti79 said:
The third line back up...I think I can take out the minus sign giving


[tex]-\|(x_n-x)\|<ε/2[/tex] and [tex]n_2 ∈ N[/tex]
Look at the definition of "norm" again.

bugatti79 said:
[tex]\|(x_n-y)\|<ε/2[/tex]...Any suggestions on what to do next?
The same thing we discussed in the other thread. This step is part of every proof that involves the definition of limits of sequences and the triangle inequality.
 
  • #12
Ok, the norm means the evaluation will be strictly positive hence it is ok to leave the minus sign inside...I think! here goes..

Let [itex]\varepsilon>0[/itex] be arbitrary. Let [itex]n_1\in\mathbb N[/itex] be such that for all [itex]n\in\mathbb N[/itex], [tex]n\geq n_1\ \Rightarrow\ \|-(x_n-x)\|<\frac{\varepsilon}{2}.[/tex] Let [itex]n_2\in\mathbb N[/itex] be such that for all [itex]n\in\mathbb N[/itex], [tex]n\geq n_2\ \Rightarrow\ \|x_n-y\|<\frac{\varepsilon}{2}.[/tex] For all [itex]n\in\mathbb N[/itex] such that [itex]n\geq \max\{n_1,n_2\}[/itex],

[tex]||(x-y)\| \leq ||-(x_n-x)||+||x_n-y|| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}< \varepsilon[/tex]...?
 
  • #13
Yes, that's correct. As you have already noticed, the proof is essentially the same as in the other thread. This is a standard proof method that you will undoubtedly have to use many times again in the future.

What i wanted you to see when I said that you should look at the definition of "norm" again, is that the definition includes the condition [itex]\|ax\|=|a|\,\|x\|[/itex]. You should use this to get rid of that minus sign.

Now there are two more things you have to make sure that you understand before this proof is complete. The first is that if t is a non-negative real number, and t<ε for all ε>0, then t=0. The second is that the only vector that has norm 0 is the zero vector.
 
  • #14
Fredrik said:
Yes, that's correct. As you have already noticed, the proof is essentially the same as in the other thread. This is a standard proof method that you will undoubtedly have to use many times again in the future.

What i wanted you to see when I said that you should look at the definition of "norm" again, is that the definition includes the condition [itex]\|ax\|=|a|\,\|x\|[/itex]. You should use this to get rid of that minus sign.

Now there are two more things you have to make sure that you understand before this proof is complete. The first is that if t is a non-negative real number, and t<ε for all ε>0, then t=0. The second is that the only vector that has norm 0 is the zero vector.

Ok, thanks! Where would t come into play in this problem or did you mean n?

Thanks in advance!
 
  • #15
more generally it is valid in haussdorf spaces because you can separate each point with two nbds which both has to contain all but finite number of the sequence hence contradictions. It will might also be enough for it to be one of those T something spaces whose definitions I don't remeber now.
 
  • #16
bugatti79 said:
Ok, thanks! Where would t come into play in this problem or did you mean n?
I'm saying that you need the following theorem:
For all t≥0, if t<ε for all ε>0, then t=0.​
It's a "for all" statement, so it doesn't matter what symbol we use. Note that it's quite common to express "for all" statements as "if" statements. For example, "for all real t, t2≥0", can be expressed as "if t is real, then t2≥0". These statements should actually both be considered abbreviated forms of the more accurate "for all t, if t is a real number, then t2>0".
 
  • #17
Sina said:
more generally it is valid in haussdorf spaces because you can separate each point with two nbds which both has to contain all but finite number of the sequence hence contradictions. It will might also be enough for it to be one of those T something spaces whose definitions I don't remeber now.

T(0,1,2,3,4) comes from separation properties. Hausdorff is T2. I suspect* the proof you gave also requires first-countability.

* somebody check!
 
  • #18
Why does it require first countability?

I can separate two points with disjoint open nbds. If the sequence converges to both, then the sequence has to have finite number of elements outside of each nbd and infinite number of elements in each nbd which is a contradiction?

By the way you can simply do the proof by showing normed spaces are haussdorf (which should be easy using triangle inequality and concept of distance coming from both linearity and the norm) -and first countability if dextercioby is right-
 
  • #19
I haven't even looked up what first countable means, but I don't think I need to. It's clear that the Hausdorff condition is sufficient, because [itex]x_n\to x[/itex] means that every open set that contains x also contains all but a finite number of terms of the sequence. If [itex]x_n\to x[/itex] and [itex]x_n\to y\neq x[/itex] in a Hausdorff space, then there are disjoint open sets U and V such that [itex]x\in U[/itex] and [itex]y\in V[/itex], and both contain all but a finite number of terms. This is a contradiction, because [itex]x_n\to x[/itex] implies that V can only contain a finite number of terms, and [itex]x_n\to y[/itex] implies that V must contain infinitely many.
 

1. What is a normed linear space?

A normed linear space is a mathematical concept that combines the properties of a vector space and a metric space. It is a set of elements (vectors) that can be added together and multiplied by scalars, and also has a defined norm function that assigns a size or length to each vector.

2. What is the purpose of a normed linear space?

The purpose of a normed linear space is to provide a mathematical framework for studying and analyzing vectors in a more general and abstract manner. It allows for the application of concepts and techniques from linear algebra and functional analysis to a wide range of problems and fields, such as physics, engineering, and economics.

3. How is the norm of a vector calculated in a normed linear space?

The norm of a vector in a normed linear space is calculated using the norm function, which is a real-valued function that assigns a non-negative value to each vector. The most common norm used is the Euclidean norm, which is calculated as the square root of the sum of the squared components of the vector.

4. What are some properties of a normed linear space?

Some properties of a normed linear space include: the triangle inequality, which states that the norm of the sum of two vectors is less than or equal to the sum of their individual norms; homogeneity, which states that the norm of a scalar multiple of a vector is equal to the absolute value of the scalar times the norm of the vector; and positive definiteness, which states that the norm of a vector is equal to 0 if and only if the vector is the zero vector.

5. How is a normed linear space different from a metric space?

A normed linear space is a special type of metric space, meaning it has all the properties of a metric space but also has the additional structure of a vector space. This means that in a normed linear space, the distance between two points (vectors) is defined by their norm, whereas in a general metric space, the distance function can take on various forms.

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