Help! Advanced Calculus

**violetsf** · Sep 24th 2007, 10:30 AM

Let K be a sequentially compact subset of R^n and suppose O is an open set of R^n that contains K. Prove that there is some positive number r such that for any point U in K, Ball (U) with radius of r is contained in O.

Hint: by contradiction.
Thank you!

**Plato** · Sep 24th 2007, 12:09 PM

Here is an outline.

Suppose NOT: $n \in Z^ + \quad \Rightarrow \quad \left( {\exists k_n } \right)\left[ {B\left( {k_n ;1/n} \right) \not\subset O} \right]$ .
Because K is sequentially compact $<br /> \left( {k_n } \right)$ has an accumulation point (limit point) is K. Call it k.
Because O covers K, $\left( {\exists \varepsilon > 0} \right)\left[ {B\left( {k;\varepsilon } \right) \subset O} \right]$ .
$\left( {\exists N \in Z^ + } \right)\left[ {\frac{1}{N} < \frac{\varepsilon }{4}} \right]\quad \& \quad \left( {\exists J \in Z^ + ,J > N} \right)\left[ {k_J \in B\left( {k;\varepsilon/2 } \right)} \right]$ .
But $\left( {\exists t \in K} \right)\left[ {t \in B\left( {k_J ;1/J} \right)\backslash O} \right]$
$d(t,k) \le d\left( {t,k_J } \right) + d\left( {k_j ,k} \right) < \frac{\varepsilon }{4} + \frac{\varepsilon }{2} < \varepsilon$ .

Do you see the contradiction there?

**violetsf** · Sep 24th 2007, 01:12 PM

Thank you very much!
It is still difficult for me, but I will follow your idea.

**Plato** · Sep 24th 2007, 01:22 PM

Originally Posted by violetsf

It is still difficult for me, but I will follow your idea.

May I suggest that you look into some textbooks for the following topics: $\varepsilon \mbox{-nets}$ , totally bounded, and/or precomapct. Each of these is close to what the given problem is asking.

**Rebesques** · Oct 4th 2007, 01:38 AM

Good job P.

Now here's another way to see this, using the properties of distance functions.

Define the distance function $d(y,\partial O)=\inf_{x\in \partial O}|x-y|, \ y\in K, \ \partial O$ denoting the boundary of O (which is nonempty as O is open). This is continuous as the norm is continuous. If we show that $\inf_{y\in K}d(y,\partial O)=d_0>0$ , then for all radii $r<d_0$ and all $y\in K$ we will have $B(y,r)\subset O$ .

Suppose on the contrary $\inf_{y\in K}d(y,\partial O)=0$ . Then there is a sequence $y_n\in K$ such that $\lim d(y_n,\partial O)=0$ . Since K is sequentially compact, the sequence $(y_n)$ has a subsequence $(y_{k_n})$ that converges to some element $y_0\in K$ .

So we get $0=\lim d(y_{k_n},\partial O)=d(y_0,\partial O)$ . This means $y_0\in \partial O$ , as $\partial O$ is closed; This contradicts the fact that $y_0\in K\subset O$ .

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