Sequence Convergence proof maybe?

**Caity** · October 26th 2008, 02:21 PM

a) Suppose that ${X_n}$ and ${Y_n}$ converge to the same point. Prove that ${X_n} - {Y_n} \rightarrow 0$ as $n \rightarrow \infty$ .

b) Show the converse is false.

Again as usual... some kind of explanation to help me understand this helps alot...

**whipflip15** · October 26th 2008, 02:57 PM

For part a) i would use the $\epsilon-\delta$ definition of limits and it should follow directly. For part b) you just need to show a counter example. Maybe look at $y=x\mbox{ and }y^2-x^2=1.$

**Caity** · October 27th 2008, 05:28 AM

ok that was how I started it... thanks for the info...

**Plato** · October 27th 2008, 07:02 AM

Originally Posted by Caity

a) Suppose that ${X_n}$ and ${Y_n}$ converge to the same point. Prove that ${X_n} - {Y_n} \rightarrow 0$ as $n \rightarrow \infty$ .
b) Show the converse is false.

Suppose that $L$ is the common limit of the two sequences.
If $\varepsilon > 0 \Rightarrow \quad \left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow \left| {x_n - L} \right| < \frac{\varepsilon } {2}\,\& \,\left| {y_n - L} \right| < \frac{\varepsilon } {2}} \right]$ .
Then observe that $\left| {x_n - y_n } \right| \leqslant \left| {x_n - L} \right| + \left| {L - y_n } \right|$ .

**Caity** · October 27th 2008, 11:01 AM

I finished part a). Thanks for the help. Now about part b). First what exactly would the converse be? I know its If X then Y becomes If Y then X. I was thinking maybe by showing something like subtraction is not commutative. Like if X - Y $\rightarrow$ 0 is not the same as Y - X $\rightarrow$ 0.

**Plato** · October 27th 2008, 12:43 PM

The converse is: Suppose that $\left( {x_n - y_n } \right) \to 0$ the $\left( {x_n } \right)$ & $\left( {y_n } \right)$ have the same limit.

**Caity** · October 28th 2008, 07:38 AM

Originally Posted by Plato

The converse is: Suppose that $\left( {x_n - y_n } \right) \to 0$ the $\left( {x_n } \right)$ & $\left( {y_n } \right)$ have the same limit.

Well I tried this Plato... but it comes out true and I'm supposed to be proving it false... If the limit of $x_n$ and $y_n$ go to the same number say 1. Then we have 1 - 1 = 0 which shows it is true.

**Plato** · October 28th 2008, 09:01 AM

Try this counterexample.
$x_n = \left( { - 1} \right)^n + \frac{1} {n}\,\& \,y_n = \left( { - 1} \right)^n - \frac{1} {n}\,$

Note: we are not given that the sequences converge, only that their difference is null.

**Moo** · October 30th 2008, 11:51 AM

Originally Posted by Plato

Note: we are not given that the sequences converge, only that their difference is null.

Is it possible to find a counterexample with the two converging ?

**Plato** · October 30th 2008, 12:39 PM

Originally Posted by Moo

Is it possible to find a counterexample with the two converging ?

The answer is no, the limit must be the same for both sequences.
Suppose that $\left( {x_n } \right) \to L,\quad \left( {y_n } \right) \to K\,\& \,\left| {\left( {x_n } \right) - \left( {y_n } \right)} \right| \to 0$ .
That means that almost all the terms of $\left( {x_n } \right)$ are ‘close’ to $L$ ; almost all the terms of $\left( {y_n } \right)$ are ‘close’ to $K$ ; and almost all the terms of $\left( {x_n } \right) - \left( {y_n } \right)$ are ‘close’ to $0$ .

Suppose $L \not= K$ then $\varepsilon = \frac{{\left| {L - K} \right|}} {4} > 0$ .
But how can almost all of the three sequences be with in an $\epsilon$ -distance of its limit?

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