# Normed Linear Space - Query

1. Dec 3, 2011

### bugatti79

Folks,

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

THanks

2. Dec 3, 2011

### micromass

Staff Emeritus
What do you mean with "convergence sequences are unique"?? I don't really understand what you're asking.

3. Dec 3, 2011

### bugatti79

if x_n is a sequence and converges to x and y then x=y. THis is what I am looking for.

4. Dec 3, 2011

### micromass

Staff Emeritus
Yes, that's true in every normed space. What are you looking for?? A proof of that?

5. Dec 3, 2011

### bugatti79

yes if it is standard text book stuff. Its not in my book. Thanks !

6. Dec 3, 2011

### micromass

Staff Emeritus
OK, I'll give you a hint on how to prove it:

$$\|x-y\|\leq \|x-x_n\|+\|x_n-y\|$$

Use the definition of convergence of a sequence now.

7. Dec 4, 2011

### bugatti79

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

$$\|x-y\|\leq \|(x-x_n)+(x_n-y)\|$$

8. Dec 4, 2011

### dextercioby

Yes, but with the = sign, since you're adding and subtracting the same number inside the norm.

9. Dec 4, 2011

### bugatti79

ok,

$$\|x-y\|\leq \|(x-x_n)+(x_n-y)\|$$

$$\|x-y\|= \|(x-x_n)+(x_n-y)\| \le \|x-x_n\|+\|x_n-y\| \le \|-(x_n-x)\|+\|x_n-y\|$$

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 $$n_1 ∈ N$$ such that

$$\|-(x_n-x)\|<ε/2$$ and $$n_2 ∈ N$$

$$\|(x_n-y)\|<ε/2$$

How I doing so far?

10. Dec 5, 2011

### bugatti79

$$n_1 ∈ N$$

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

$$-\|(x_n-x)\|<ε/2$$ and $$n_2 ∈ N$$

$$\|(x_n-y)\|<ε/2$$...Any suggestions on what to do next?

11. Dec 5, 2011

### Fredrik

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

Look at the definition of "norm" again.

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. Dec 5, 2011

### bugatti79

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

$$||(x-y)\| \leq ||-(x_n-x)||+||x_n-y|| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}< \varepsilon$$....?

13. Dec 5, 2011

### Fredrik

Staff Emeritus
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 $\|ax\|=|a|\,\|x\|$. 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. Dec 5, 2011

### bugatti79

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

15. Dec 5, 2011

### Sina

more generally it is valid in haussdorf spaces because you can seperate 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. Dec 5, 2011

### Fredrik

Staff Emeritus
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. Dec 5, 2011

### dextercioby

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. Dec 7, 2011

### Sina

Why does it require first countability?

I can seperate 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. Dec 7, 2011

### Fredrik

Staff Emeritus
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 $x_n\to x$ means that every open set that contains x also contains all but a finite number of terms of the sequence. If $x_n\to x$ and $x_n\to y\neq x$ in a Hausdorff space, then there are disjoint open sets U and V such that $x\in U$ and $y\in V$, and both contain all but a finite number of terms. This is a contradiction, because $x_n\to x$ implies that V can only contain a finite number of terms, and $x_n\to y$ implies that V must contain infinitely many.