# Help! Advanced Calculus

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• Sep 24th 2007, 10:30 AM
violetsf
Help! Advanced Calculus
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!
• Sep 24th 2007, 12:09 PM
Plato
Here is an outline.

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

Do you see the contradiction there?
• Sep 24th 2007, 01:12 PM
violetsf
Thank you very much!
It is still difficult for me, but I will follow your idea.
• Sep 24th 2007, 01:22 PM
Plato
Quote:

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: $\displaystyle \varepsilon \mbox{-nets}$, totally bounded, and/or precomapct. Each of these is close to what the given problem is asking.
• Oct 4th 2007, 01:38 AM
Rebesques
Good job P. :)

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

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

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

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