Homework Help: Show that this is a metric.

1. Feb 1, 2014

bobby2k

1. The problem statement, all variables and given/known data
A sequence $\{x_{n}\}$ of real numbers is called bounded if there is a number M such that |$x_{n}$| ≤M for all n. Let X be the set of all bounded sequences, show that
$d(\{x_{n}\},\{y_{n}\})=sup \{|x_{n}-y_{n}| :n \in N \}$ is a metric on X.

The only part I am struggling with is the transivity part. As I see it, I have to show that:

$sup \{|x_{n}-y_{n}| :n \in N \}≤sup \{|x_{n}-z_{n}| :n \in N \}+sup \{|z_{n}-y_{n}| :n \in N \}$

Do you guys have any tips on how to show this? The problem is that I can not be sure that I have an n where I get the sup value, and even if I did, this n might be different for the three parts.

Last edited: Feb 1, 2014
2. Feb 1, 2014

LCKurtz

Start with $|x_n-y_n|\le |x_n-z_n|+|z_n-y_n|$ and take the sup of both sides. Then compare what that gives with the right side of your expression.

3. Feb 2, 2014

bobby2k

It became a little messy, can I simplify?

$|x_{n}-y_{n}|≤ |x_{n}-z_{n}|+|z_{n}-y_{n}| \forall n$ (1)
$|x_{n}-z_{n}| ≤ sup\{|x_{n}-z_{n}| : n \in N \} \forall n$ (2)
$|z_{n}-y_{n}| ≤ sup\{|z_{n}-y_{n}| : n \in N \} \forall n$ (3)

So if I put 2 and 3 in 1 I get:
$|x_{n}-y_{n}|≤ sup\{|x_{n}-z_{n}| : n \in N \} + sup\{|z_{n}-y_{n}| : n \in N \} \forall n$ (4)

Hence $sup\{|x_{n}-z_{n}| : n \in N \} + sup\{|z_{n}-y_{n}| : n \in N \}$ is an upper bound for $\{|x_{n}-y_{n}| :n\in N\}$. So it must also be bigger than the least upper bound for this set. Hence:
$sup\{|x_{n}-y_{n}| : n \in N \} ≤ sup\{|x_{n}-z_{n}| : n \in N \} +sup\{|z_{n}-y_{n}| : n \in N \}$

Is this the simplest way do it, or is it a simpler way?

Last edited: Feb 2, 2014
4. Feb 2, 2014

LCKurtz

There may be slightly nicer ways to write it up but I see nothing wrong with your argument.