Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,||\cdot||)$ be a complete normed space. Let $F_1, F_2, F_3,\ldots\subseteq X$ be closed, non-empty subsets of $X$.

Assume that $F_1 \supseteq F_2\supseteq F_3\supseteq \cdots$

and that $\sup_{x,y\in F_n}||x-y||\to0$.

Show that the intersection of all $F_n$ is non-empty.

All I can possibly think to go on is that we need to show that there is some $f\in X$ such that $f\in F_1,F_2,F_3,\ldots$

Any pointers in the right direction?

share|improve this question
2  
For each $n \in \mathbb{N}$ pick $x_n \in F_n$, then use your assumption to show the sequence is Cauchy and conclude by proving that the limit is in the intersection (notice that if $n \ge m$, then $x_n \in F_m$, and use the fact that $F_m$ is closed). –  Joel Cohen Mar 1 '12 at 0:28
1  
This seems like it should work in any complete metric space. Here are two hints: (1) for each $n$, what does the fact that $F_n$ is non-empty tell you? (2) how are we going to use completeness in proving the result? –  user16299 Mar 1 '12 at 0:29
    
Ah, I see Joel beat me to the punch. –  user16299 Mar 1 '12 at 0:29

1 Answer 1

up vote 2 down vote accepted

If we let $x_n \in F_n$, then you can check that $x_1,x_2,\ldots $ is a Cauchy sequence converging (since $X$ is complete) to $x \in X$.

Now fix an $i$. If $j>i$, then $x_j \in F_j \subset F_i$, so since $F_i$ is closed, it follows that $x \in F_i$. Since $x \in F_i$ for all $i$, it follows that $x \in \cap_i F_i$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.