# Math Help - 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.

Thank you!

2. 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 $
\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?

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

4. 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.

5. 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 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 $y_0\in K\subset O$.