# Convergent sequences in Cartesian product of vector spaces

1. Oct 2, 2012

### Greger

If A and B are vector spaces over ℝ or ℂ show that a sequence (a_n, b_n) in A×B converges to (a,b) in A×B only if a_n converges to a in A and b_n converges to b in B as n tends to infinity.

To me this statement sounds pretty intuitive but I have been having trouble actually proving it properly.

My first attempt was to assume that a_n converges to a in A and b_n converges to b in B then it was kind of easy to see that (a_n, b_n) converges to (a,b) but showing that the converse is true seems to be a bit trickier.

To me it seems like if you have a sequence (a_n, b_n) where a_n is in A and b_n is in B that it is 'obvious' that it converges to (a,b) only if if the individual sequences converge in their space. I mean like, if it didn't converge in it's space, how could it converge in the Cartesian product?

Does anyone have any ideas on finishing off this proof? (assuming that I started with the correct idea)

2. Oct 2, 2012

### micromass

Staff Emeritus
Vector spaces?? How can you talk about convergence in just vector spaces?

3. Oct 2, 2012

### Greger

Do you mean that it should be a normed vector space?

Like in this case |(a,b)| = |a|_A + |b|_B for a in A and b in B

Or?

4. Oct 2, 2012

### micromass

Staff Emeritus
Yes, A and B should be normed vector spaces (or metric spaces more generally). And you should make a norm on AxB as well. Your proposal of the norm

$$\|(a,b)\|=\|a\|_A+\|b\|_B$$

is a good one.

Now, can you write out what it means that the sequence $(a_n,b_n)$ converges in AxB? And what it means that $a_n$ converges in A and that $b_n$ converges in B?

5. Oct 2, 2012

### Greger

For the cartesian product,

For some ε>0 there exists an N such that |(a_n,b_n) - (a,b)|<ε whenever n>N

|(a_n-a,b_n-b)| = |a_n-a|_A + |b_n-b|_B

For individual spaces

For some ε>0 there exists an N such that |a_n - a|_A<ε/2 whenever n>N
(same definition for b_n)

and if you add them together you will get

|a_n - a|_A + |b_n - b|_A<ε/2+ε/2=ε

but then to prove it don't you also need to show that the converse of the statement is not true?

Is it enough to say that

For some ε>0 there exists an N such that |(a_n,b_n) - (a,b)|<ε whenever n>N

|(a_n-a,b_n-b)| = |a_n-a|_A + |b_n-b|_B<ε

but if there does not exist an N such that |a_n-a|_A <ε/2 for n>N then it doesn't converge ?

6. Oct 2, 2012

### micromass

Staff Emeritus
Same for the $b_n$, but the N there might be different than the N for the $a_n$. So how do you solve that?

What do you mean?

7. Oct 2, 2012

### Greger

Sorry I editing my post a bit to make it make more sense haha,

If the N are different then for the a case i'll write it as N_a, then you need N = max(N_a, N_b) right?

8. Oct 2, 2012

### micromass

Staff Emeritus
Are you trying to show now that if $(a_n,b_n)$ converges, then $a_n$ converges. To prove this you have to find for all $\varepsilon>0$ an N such that for n>N holds that $|a_n-a|_A<\varepsilon$. Ho< would you find that N?

Correct.

9. Oct 2, 2012

### Greger

Wouldnt that N be the same N required for (a_n,b_n) to converge?

10. Oct 2, 2012

### micromass

Staff Emeritus
Yes, try to use that N. So you know that

$$\|(a_n,b_n)-(a,b)\|=\|a_n-a\|_A+\|b_n-b\|_B<\varepsilon$$

and you must prove that

$$\|a_n-a\|_A<\varepsilon$$

Can you do this?