Closed subset of R^n has an element of minimal norm

[kingwinner]

Homework Statement

a) Let V be a normed vector space. Then show that (by the triangle inequality) the function f(x)=||x|| is a Lipschitz function from V into [0,∞). In particular, f is uniformly continuous on V.

b) Show that a closed subset F of contains an element of minimal norm, that is, there is an x E F such that ||x||≤||y|| for all y E F. (here ||x|| refers to the usual Euclidean norm).
(hint: F may not be compact, so work on a suitable compact subset of F.)

Homework Equations

The Attempt at a Solution

I proved part a, but I really have no idea how to do part b. Why do we need compactness, and what is the suitable compact subset of F?

I hope someone can help me out! Thank you!

 

[lanedance]

just ideas that will need some working but hope they help

you want to take a compact subset of F, is a closed an bounded subset enough to be compact (it is in Rⁿ - Hiene Borel)?

you could pick any element of F, then use it to set your bound, and show elements outside the bounded set have a larger norm

now you have a compact subset of a metric space and i seem to remember something along the lines of a continuous real valued function always takes a maximal & minimal value on a compact set...

 

[kingwinner]

If we intersect F with a compact set K, will F intersect K stilll be compact? Why or why not?
Also, should we choose K to be very small? How can we formalize this choice of K? (i.e. what is the so-called suitable compact subset of F mentioned in the hint?)


"...and show elements outside the bounded set have a larger norm"
How??

 

[JSuarez]

Consider the closed balls of radius r, centred at the origin:

$$B\left(0,r\right)=\left\{x \in V:\left\|x\right\|\leq r\right\}$$

There must be one that intersects F. How do you go on from there?

 

[kingwinner]

Why does it have to intersect F?

And how can we show elements outside the bounded set have a larger norm? I don't get this idea at all. How is this possible? Can someone please explain this in greater detail?

Thanks a lot!

 

[Office_Shredder]

By definition, elements outside the ball of radius r have a larger norm than elements inside a ball of radius r. Look at how Suarez described the set

 

[JSuarez]

For sufficiently large r, it must intersect F; otherwise, F would be empty.

 

[kingwinner]

What would go wrong if F were empty??

 

[JSuarez]

It wouldn't have an element with minimum norm; in fact, the original statement of your question should be "a closed nonempty subset F contains an element of minumum norm". As the empty set is closed, the statement is false without that condition.