# inf and sup of a compact subset of R

• Feb 28th 2010, 08:52 PM
inthequestofproofs
inf and sup of a compact subset of R
Let K be a compact set
I have to show that for some a, b in K, we can show |a-p| = inf{|x-p|: x in K} and |b-p|=sup{|x-p|: x in K}

My proof
Since K is compact, K is closed and bounded. Since K is bounded, it is bounded above and below. that is, there is a B in R s.t. B>=x for all x in K. B is the least upper bound. Similarly, there is a A in R s.t. A <= x for all x in K. A is the greatest lower bound.

Let p<k for all k in K. Then choose a = min {values in K}. Then a inf|x-p| = |a-p|

Let p>k for all k in K. Then choose b=man {values in K}. Then sup|x-p|=|b-p|

I don't think my last part is right. Please provide me with some feedback
• Feb 28th 2010, 09:00 PM
Drexel28
Quote:

Originally Posted by inthequestofproofs
Let K be a compact set
I have to show that for some a, b in K, we can show |a-p| = inf{|x-p|: x in K} and |b-p|=sup{|x-p|: x in K}

My proof
Since K is compact, K is closed and bounded. Since K is bounded, it is bounded above and below. that is, there is a B in R s.t. B>=x for all x in K. B is the least upper bound. Similarly, there is a A in R s.t. A <= x for all x in K. A is the greatest lower bound.

Let p<k for all k in K. Then choose a = min {values in K}. Then a inf|x-p| = |a-p|

Let p>k for all k in K. Then choose b=man {values in K}. Then sup|x-p|=|b-p|

I don't think my last part is right. Please provide me with some feedback

I have no idea what this means, but I would assume we're dealing with some metric space $K$ is some compact metric space, and we have some $x\in X-K$. Show that $d(x,K)=d(x,k)$ for some $k\in K$.

Note that $\rho:K\mapsto\mathbb{R}$ given by $k\mapsto d(k,x)$ is a continuous mapping, and since $K$ is compact we have that $\rho$ assumes a minimum on $K$. The conclusion follows.
• Feb 28th 2010, 09:03 PM
inthequestofproofs
compact sets
Thank you! We haven't covered metric sets. This is a tough proof, but your insight might provide some clarity
• Feb 28th 2010, 09:04 PM
Drexel28
Quote:

Originally Posted by inthequestofproofs
Thank you! We haven't covered metric sets. This is a tough proof, but your insight might provide some clarity

Oh! I'm sorry, I didn't notice it said $\mathbb{R}$. The same concept applies, just replace the metric space with $\mathbb{R}$.