Can anybody help me to understand this proof?

1. Sep 30, 2012

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 $(x_n)$ in $F_n$. By the Bolzano-Weierstrass theorem 13.5, a subsequence $(x_n_m})^{\infty}_{m=1}$ of $(x_n)$ converges to some element $x_0$ in $R^k$. To show $x_0 \in F$, it suffices to show $x_0 \in F_n_0$ with $n_0$ fixed.

Why does it suffice to show that?

If $m \geq n_0$, then $n_m \geq n_0$, so $x_n_m \in F_n_m \subseteq F_n_0$.

But what if m<n_0?

Hence the sequence ${x_n_m}^{\infty}_{m=1}$ consists of points in $F_n_0$ and converges to $x_0$. Thus $x_0$ 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.)

Last edited: Sep 30, 2012
2. Sep 30, 2012

DonAntonio

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 [tex] option here but itex enclosed in square parentheses. Please do fix your post.

DonAntonio

3. Sep 30, 2012

Artusartos

But I don't know how to use itex...??

4. Sep 30, 2012

Einj

Inside  you have to write in latex language.

5. Sep 30, 2012

Artusartos

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

6. Oct 1, 2012

micromass

Staff Emeritus
Last edited: Oct 1, 2012
7. Oct 2, 2012

Erland

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....

8. Oct 2, 2012

Artusartos

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

9. Oct 2, 2012

Erland

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.