null sequences.

Printable View

May 6th 2011, 10:37 AM
abhishekkgp

null sequences.

Prove or disprove:
If http://latex.codecogs.com/png.latex?(x_ny_n) is a null sequence then one or both of http://latex.codecogs.com/png.latex?(x_n), \, (y_n) is(are) null.

My attempt:
if http://latex.codecogs.com/png.latex?..., and \, (y_n) both are null then certainly http://latex.codecogs.com/png.latex?(x_ny_n) is null, no problem with that.

I assumed that http://latex.codecogs.com/png.latex?(x_n) is not null, http://latex.codecogs.com/png.latex?(y_n) is not null and as given, http://latex.codecogs.com/png.latex?(x_ny_n) is null. I tried to bring a contradiction since i have a feeling that the statement in the question is true.

since its assumed that http://latex.codecogs.com/png.latex?(x_n) is not null so:
there exists http://latex.codecogs.com/png.latex?\epsilon_1>0 such that for all http://latex.codecogs.com/png.latex?N there exists http://latex.codecogs.com/png.latex?n \geq N such that http://latex.codecogs.com/png.latex?|x_n|>\epsilon_1

similarly:
there exists http://latex.codecogs.com/png.latex?\epsilon_2>0 such that for all http://latex.codecogs.com/png.latex?N there exists http://latex.codecogs.com/png.latex?n \geq N such that http://latex.codecogs.com/png.latex?|y_n|>\epsilon_2

and since http://latex.codecogs.com/png.latex?(x_ny_n) is null:
there exists http://latex.codecogs.com/png.latex?N such that http://latex.codecogs.com/png.latex?...n_1 \epsilon_2

now since i can't say that http://latex.codecogs.com/png.latex?..._n|>\epsilon_2 occur at the same http://latex.codecogs.com/png.latex?n i can't find a contradiction.

someone help.
May 6th 2011, 10:42 AM
Plato

Quote:

Originally Posted by abhishekkgp

Prove or disprove:
If http://latex.codecogs.com/png.latex?(x_ny_n) is a null sequence then one or both of http://latex.codecogs.com/png.latex?(x_n), \, (y_n) is(are) null.

Assuming both converge, suppose that neither is a null sequence.
Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$ .
What can you say about $(x_ny_n)\to~ ?$
What does that tell you.
May 6th 2011, 10:51 AM
abhishekkgp

Quote:

Originally Posted by Plato

Assuming both converge, suppose that neither is a null sequence.
Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$ .
What can you say about $(x_ny_n)\to~ ?$
What does that tell you.

probably http://latex.codecogs.com/png.latex?...rrow XY \neq 0 and probably null sequences converge to 0 and hence a contradiction. But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess.
May 6th 2011, 11:01 AM
Plato

Quote:

Originally Posted by abhishekkgp

But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess.

How very odd. I know Berberian's book on measure theory and find it odd.
Pray tell then what is a null sequence?
May 6th 2011, 11:05 AM
abhishekkgp

Quote:

Originally Posted by Plato

How very odd. I know Berberian's book on measure theory and find it odd.
Pray tell then what is a null sequence?

a sequence http://latex.codecogs.com/png.latex?(x_n) in http://latex.codecogs.com/png.latex?\mathbb{R} is said to be null if:
for all http://latex.codecogs.com/png.latex?\epsilon>0 there exists http://latex.codecogs.com/png.latex?N such that http://latex.codecogs.com/png.latex?...|x_n|<\epsilon
May 6th 2011, 11:13 AM
Plato

Quote:

Originally Posted by abhishekkgp

a sequence http://latex.codecogs.com/png.latex?(x_n) in http://latex.codecogs.com/png.latex?\mathbb{R} is said to be null if:
for all http://latex.codecogs.com/png.latex?\epsilon>0 there exists http://latex.codecogs.com/png.latex?N such that http://latex.codecogs.com/png.latex?...|x_n|<\epsilon

OK go back to the OP.
Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$ .
Make $|x_ny_n|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le |x_n||y_n|=|x_ny_n|<\varepsilon$ .
That is a contradiction.
May 6th 2011, 11:20 AM
abhishekkgp

Quote:

Originally Posted by Plato

OK go back to the OP.
Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$ .
Make $|x_ny_n|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le |x_n||y_n|=|x_ny_n|<\varepsilon$ .
That is a contradiction.

the problem is that i can't( or yet not able to) say that there exists an http://latex.codecogs.com/png.latex?n such that http://latex.codecogs.com/png.latex?...n_1 \epsilon_2. I can only say that http://latex.codecogs.com/png.latex?...n_1 \epsilon_2 and that's why i can't contradict anything.
May 6th 2011, 03:44 PM
Plato

Quote:

Originally Posted by abhishekkgp

the problem is that i can't( or yet not able to) say that there exists an http://latex.codecogs.com/png.latex?n such that http://latex.codecogs.com/png.latex?...n_1 \epsilon_2. I can only say that http://latex.codecogs.com/png.latex?...n_1 \epsilon_2 and that's why i can't contradict anything.

I think that I misread the OP.
Consider this example.
$x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$
What is $x_ny_n~?$
May 6th 2011, 09:58 PM
abhishekkgp

Quote:

Originally Posted by Plato

I think that I misread the OP.
Consider this example.
$x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$
What is $x_ny_n~?$

so it can happen that if we have $(x_n)$ not null, $(y_n)$ not null, we still have $(x_ny_n)$ as null!!

thank you for this.