Is the Metric Space (X,d) Separable and Compact?

[CalTech>MIT]

Homework Statement

X={x | x_n E R | 0$$\leq$$ x $$\leq$$ 1}
d(x,y)= $$\Sigma$$_n=1^infinity |x_n - y_n|*2^-j
Show:
1. (X,d) is a metric space
2. (X,d) is separable
3. (X,d) is compact

Homework Equations

n/a

The Attempt at a Solution

Here we go.
number 1.
Show that d(x,y)=d(y,x):
$$\Sigma$$_n=1^infinity |x_n - y_n|*2^-j = $$\Sigma$$_n=1^infinity |y_n - x_n|*2^-j

Show that d(x,x)=0:
$$\Sigma$$_n=1^infinity |x_n - x_n|*2^-j = $$\Sigma$$_n=1^infinity 0*2^-j = 0

Show d(x,y)$$\leq$$d(x,z)+d(z,y):
$$\Sigma$$_n=1^infinity |x_n - z_n|*2^-j + $$\Sigma$$_n=1^infinity |z_n - y_n|*2^-j

 

[micromass]

Hello, welcome to physicsforums!

So you've proven that it is a metric space. That's good!
Now it suffices to show 3 (since every compact metric space is separable).

So take a sequence in X, we will show that it has a convergent subsequence. The sequence in X has the form
$$(x_{1,1}, x_{1,2}, x_{1,3}, x_{1,4},...)$$
$$(x_{2,1}, x_{2,2}, x_{2,3}, x_{2,4},...)$$
$$(x_{3,1}, x_{3,2}, x_{3,3}, x_{3,4},...)$$
$$(x_{4,1}, x_{4,2}, x_{4,3}, x_{4,4},...)$$

The first vertical sequence has a convergent subsequence, say $$x_{k^1_n,1}\rightarrow x_1$$. Now consider the sequence $$x_{k^1_n,2}$$, this has a convergent subsequence, say $$x_{k^2_n,2}\rightarrow x_2$$. And so on and so on.

Now I claim that the above sequence has a convergent subsequence which converges to $$(x_1,x_2,x_3,...)$$. Can you see which one?

 

[CalTech>MIT]

I believe it'd be: (x_1,1, x_2,2, x_3,3, ...)?

 

[micromass]

That makes no sense... What you wrote is only one element. And it's not even an element of the sequence...

You'll need to find infinitly many "sequences"...

 

[micromass]

I'll try to give the subsequence. But it's not easy, try to visualize it.

So the construction is as follows:
Write our original sequence once more, this is

$$(x_{1,1},x_{1,2},x_{1,3},x_{1,4},...)$$
$$(x_{2,1},x_{2,2},x_{2,3},x_{2,4},...)$$
$$(x_{3,1},x_{3,2},x_{3,3},x_{3,4},...)$$
$$(x_{4,1},x_{4,2},x_{4,3},x_{4,4},...)$$

The first vertical sequence, i.e. $$(x_{n,1})_n$$ has a convergent subsequence $$(x_{k^1_n,1})_n$$ which converges to $$x_1$$
The sequence $$(x_{k^1_n,2})_n$$ has a convergent subsequence which converges to $$x_2$$.
(So we take subsequences of subsequences)

Now take the first element of the first subsequence, i.e. take $$x_{k^1_1,1}$$ and take the corresponding element: $$(x_{k^1_1,1},x_{k^1_1,2},x_{k^1_1,3},...)$$.
Now take the first element of the second subsequence, i.e. take $$x_{k^2_1,2}$$ and take the corresponding element: $$(x_{k^2_1,1},x_{k^2_1,2},x_{k^2_1,3},...)$$.
Keep repeating this, we obtain the following sequence:
$$(x_{k^1_1,1},x_{k^1_1,2},x_{k^1_1,3},...)$$.
$$(x_{k^2_1,1},x_{k^2_1,2},x_{k^2_1,3},...)$$.
$$(x_{k^3_1,1},x_{k^3_1,2},x_{k^3_1,3},...)$$.
$$(x_{k^4_1,1},x_{k^4_1,2},x_{k^4_1,3},...)$$.

Try to show that this sequence converges...

 

[CalTech>MIT]

How exactly would you prove this using the original equation?

 

[micromass]

It's really hard to explain without excessive notation I'd like it better if you could come up with it, instead of me saying how the proof goes...