# Can anybody help me to understand this proof?

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:

## Answers and Replies

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.)

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

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

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

Inside  you have to write in latex language.

Inside  you have to write in latex language.

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

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

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?
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 $x_n_m \in F_n_m \subseteq F_n_0$.

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....

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