Can anybody help me to understand this proof?

[Artusartos]

Let (F_n) be a decreasing sequence [i.e., F_1 contains F2 which contains F_3...etc] of closed bounded nonempty sets in R^k. Then F = \cap^{\infty}_{m=1} F_n is also closed, bounded and nonempty.

The proof (from our textbook) says:

Clearly F is closed and bounded. It is the nonemptiness that needs proving! For each n, select an element (<b>x</b>_n) in F_n. By the Bolzano-Weierstrass theorem 13.5, a subsequence (<b>x</b>_n_m})^{\infty}_{m=1} of (<b>x</b>_n) converges to some element <b>x</b>_0 in R^k. To show <b>x</b>_0 \in F, it suffices to show <b>x</b>_0 \in F_n_0 with n_0 fixed.

Why does it suffice to show that?

If m \geq n_0, then n_m \geq n_0, so <b>x</b>_n_m \in F_n_m \subseteq F_n_0.

But what if m<n_0?

Hence the sequence {<b>x</b>_n_m}^{\infty}_{m=1} consists of points in F_n_0 and converges to <b>x</b>_0. Thus <b>x</b>_0 belongs to F_n_0 by (b) of proposition 13.9 (which says “The set E is closed if and only if it contains the limit ofevery convergent sequence of points in E.)

Thanks in advance

 

[DonAntonio]

Let (F_n) be a decreasing sequence [i.e., F_1 contains F2 which contains F_3...etc] of closed bounded nonempty sets in R^k. Then F = \cap^{\infty}_{m=1} F_n is also closed, bounded and nonempty.

The proof (from our textbook) says:

Clearly F is closed and bounded. It is the nonemptiness that needs proving! For each n, select an element (<b>x</b>_n) in F_n. By the Bolzano-Weierstrass theorem 13.5, a subsequence (<b>x</b>_n_m})^{\infty}_{m=1} of (<b>x</b>_n) converges to some element <b>x</b>_0 in R^k. To show <b>x</b>_0 \in F, it suffices to show <b>x</b>_0 \in F_n_0 with n_0 fixed.

Why does it suffice to show that?

If m \geq n_0, then n_m \geq n_0, so <b>x</b>_n_m \in F_n_m \subseteq F_n_0.

But what if m<n_0?

Hence the sequence {<b>x</b>_n_m}^{\infty}_{m=1} consists of points in F_n_0 and converges to <b>x</b>_0. Thus <b>x</b>_0 belongs to F_n_0 by (b) of proposition 13.9 (which says “The set E is closed if and only if it contains the limit ofevery convergent sequence of points in E.)

Thanks in advance

As you could see, your post is very hard to understand. This wouldn't have happened had you used the "Preview Post" option before posting it.

There is no option here but itex enclosed in square parentheses. Please do fix your post.<br /> <br /> DonAntonio

 

[Artusartos]

As you could see, your post is very hard to understand. This wouldn't have happened had you used the "Preview Post" option before posting it.

There is no option here but itex enclosed in square parentheses. Please do fix your post.<br /> <br /> DonAntonio

<br /> <br /> But I don&#039;t know how to use itex...??

 

[Einj]

Inside you have to write in latex language.

 

[Artusartos]

Inside you have to write in latex language.

But I did do that...it still didn't change...

 

[micromass]

Here is a latex guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

You can do something like [noparse]<b>x</b>[/noparse]. So you can't use [noparse][/noparse] while using latex.

 

[Erland]

I agree that the post needs to be fixed, but I think I understand it as it is:

To show <b>x</b>_0 \in F, it suffices to show <b>x</b>_0 \in F_n_0 with n_0 fixed.

Why does it suffice to show that?

i agree that this is confusing. It is poorly expressed. The author must mean something like this:

To show x_0 in F, it suffices to show that for each fixed n_0: x_0 in F_n_0.

If m \geq n_0, then n_m \geq n_0, so <b>x</b>_n_m \in F_n_m \subseteq F_n_0.

But what if m<n_0?

This doesn't matter. The convergence behaviour of a sequence does not depend upon its first finite number of elements. If x1,x2,x3,x4,x5... converges to x, so does e.g. y1,y2,y3,x4,x5...

 

[Artusartos]

I agree that the post needs to be fixed, but I think I understand it as it is:



i agree that this is confusing. It is poorly expressed. The author must mean something like this:

To show x_0 in F, it suffices to show that for each fixed n_0: x_0 in F_n_0.

This doesn't matter. The convergence behaviour of a sequence does not depend upon its first finite number of elements. If x1,x2,x3,x4,x5... converges to x, so does e.g. y1,y2,y3,x4,x5...

Thanks for answering, but I'm not sure if I understood the first answer. What exactly does "for each fixed n_0" mean?

 

[Erland]

Thanks for answering, but I'm not sure if I understood the first answer. What exactly does "for each fixed n_0" mean?

That this should be proved for every number n_0 (which will be fixed in the succeeding part of the proof), not just that there is one n_0 for which this holds.