Re: Distance between a closed and a compact set is positive.

Cc:

In article <1146159716.333637.184940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
alencar1980 <alencar1980@xxxxxxxxxx> wrote:

I'm stuck in the following problem.

Let V be a normed spaca. If C is a closed subset of V and K a compact
subset of V such that C \intersection K = \emptyset then there exist a
r>0 such that
||c-k||>= r for all c \in C and k \in K.

My reasoning till know is as follows:

Let x \in K.

As C is closed and C \intersection K = \emptyset there exist a

r_{x} >0

such B_{r_{x}}(x) (open ball centered at x with radii r_{x}).

such that B_{r_{x}}(x)/\ C = empty... presumably...

Let H = \union_{x\in K} B_{r_{x}}(x)

As K is compact and H is an open cover of it there exist x_{i} \in K
and r_{x_i}, i=1, \ldots, p
such that

\union_{i=1, 2, ..., p} B_{r_{x_i}}(x_i)

covers K...

Am I in the right direction?

Yes, but you need to work a bit harder.

I'm going to change your notation slightly to avoid so many
subindices. Let

B(z,r) be the open ball with center z and radius r.

As above, for every x there exists r(x)>0 such that B(x,r(x))/\C is
empty. Define s(x) = r(x)/2. As above, the balls B(x,s(x)) cover K, so
there exists a finite subcover corresponding to poinst x1, ..., xp.

Let s = min{s(x1), ..., s(xp)}.

Now, for any y in K, there exists j such that the distance from y to
xj is strictly smaller than s(xj). And for every c in C and j=1,...,p, the
distance from xj to c is strictly larger than 2s(xj).

I claim that s will work.

To see this, let y in K and c in C. Then there exists j such that
d(y,xj)<s(xj). We have

d(xj,c) <= d(xj,y)+d(y,c)

We also know that d(xj,y) < s(xj) and d(xj,c) > 2s(xj). So


d(y,c) >= d(xj,c) - d(xj,y)
> 2*s(xj) - d(xj,y)
> 2*s(xj) - s(xj)
= s(xj)
>= s.

This holds for all y in K and all c in C, which proves the result.


How can I conclude the proof?



Here is another approach.

For every x in X, define the distance from x to C, d(x,C) to be

d(x,C) = inf{ d(x,c) : c in C}.

If d(x,C)=0 then for every e>0 there exists y_e in
C such that d(x,y_e)<e, hence x is a limit point of C, and since C is
closed, that means that x is in C. Thus, d(x,C)=0 if and only if x is
in C.

Now define a function f:K->[0,infty) by f(k) = d(k,C). It is not hard
to verify that this function is continuous. Since the domain is
compact, the function has a minimum; say the minimum is r, obtained at
k. Since k is not in C by hypothesis, r>0. And since it is a minimu,
for all x in K, d(x,C) >= r. And this means that for every c in C,
d(x,c)>= d(x,C) >= r.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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