# Compact metric

• May 13th 2008, 12:47 AM
namelessguy
Compact metric
I know I have to check 3 properties of a metric space, but I don't know how to approach it here. Could somebody help?
Let M be the set of all bounded sequences ( $x_n$) of real numbers:
1) Show that d( ( $x_n$), ( $y_n$) )= sup {absolute value of $x_n$ - $y_n$: n in $N$ } defines a metric on M
2) Prove that (M,d) is not compact.
• May 13th 2008, 07:53 AM
Plato
Quote:

Originally Posted by namelessguy
I know I have to check 3 properties of a metric space, but I don't know how to approach it here. Could somebody help?
Let M be the set of all bounded sequences ( $x_n$) of real numbers:
1) Show that d( ( $x_n$), ( $y_n$) )= sup {absolute value of $x_n$ - $y_n$: n in $N$ } defines a metric on M
2) Prove that (M,d) is not compact.

Where in part 1 are you having trouble? That is an exercise in absolute value.

For part 2, consider the set of all sequences $\left\{ {x_n } \right\}$ in which there are zeros everywhere except a 1 in the nth position.
• May 13th 2008, 02:31 PM
namelessguy
Quote:

Originally Posted by Plato
Where in part 1 are you having trouble? That is an exercise in absolute value.

For part 2, consider the set of all sequences $\left\{ {x_n } \right\}$ in which there are zeros everywhere except a 1 in the nth position.

The way the metric is defined in supremum term threw me off. I didn't know if I can prove this just as it's with absolute value.Here are my arguments:
a) Since ( $x_n$) and ( $y_n$) are bounded and monotonic in $R$, they both converge.
d( $x_n$, $y_n$)=sup abs ( $x_n$ - $y_n$) greater than or equal to 0, for all $x_n$, $y_n$ since absolute value of $x_n$ - $y_n$ is greater than or equal to 0, and the least upper bound must also be greater than or equal to 0.
Now for the supremum to be equal 0, the absolute value of $x_n$ - $y_n$ must be 0, which means $x_n$ = $y_n$
b) Similarly, sup of abs $x_n$ - $y_n$ = sup abs $y_n$ - $x_n$
c) I was thinking of using the triangle inequality for this one, but I don't know if it works analogously as it works for absolute value without the supremum.

For part 2) I'm thinking about the closed and bounded properties of the set you mention, but I didn't get anywhere.
• May 13th 2008, 02:45 PM
Plato
For the last question look back at the set of sequences I suggested.
Here are the first three terms: $x_1 = \left( {1,0,0,0, \cdots } \right),\quad x_2 = \left( {0,1,0,0,0 \cdots } \right),\quad x_3 = \left( {0,0,1,0,0, \cdots } \right)$.
Is it clear to you that $d\left( {x_1 ,x_2 } \right) = 1,\quad d\left( {x_1 ,x_3 } \right) = 1,\quad d\left( {x_3 ,x_2 } \right) = 1$?
Every term in that sequence is one unit from any other term in the set.
But that is a bounded, infinite set. Can it have a limit point?
• May 14th 2008, 01:05 AM
namelessguy
Quote:

Originally Posted by Plato
For the last question look back at the set of sequences I suggested.
Here are the first three terms: $x_1 = \left( {1,0,0,0, \cdots } \right),\quad x_2 = \left( {0,1,0,0,0 \cdots } \right),\quad x_3 = \left( {0,0,1,0,0, \cdots } \right)$.
Is it clear to you that $d\left( {x_1 ,x_2 } \right) = 1,\quad d\left( {x_1 ,x_3 } \right) = 1,\quad d\left( {x_3 ,x_2 } \right) = 1$?
Every term in that sequence is one unit from any other term in the set.
But that is a bounded, infinite set. Can it have a limit point?

It is clear to me that every term in that sequence is one unit from another term in the set. The set doesn't have a limit point since the term get more distant from the first term infinitely, I think. If it doesn't have a limit point, then it can't be closed.
Are my arguments in part 1) correct? If not, can you help by pointing out some mistake I made?