Proving Existence of Min. Distance in ##S## from ##p_0##

[Lee33]

Homework Statement

Prove that if $S$ is a nonempty closed subset of $E^n$ and $p_0\in E^n$ then $\min\{d(p_0,p):p\in S\}$ exists.

2. The attempt at a solution

If $p_0$ was in $S$ why would $\min\{d(p_0,p):p\in S\} = 0?$ Is it just because it is the minimum? How about if $p_0 \in S$ then what will $\max\{d(p_0,p):p\in S\}$ be?

 

[Dick]

Homework Statement

Prove that if $S$ is a nonempty closed subset of $E^n$ and $p_0\in E^n$ then $\min\{d(p_0,p):p\in S\}$ exists.

2. The attempt at a solution

If $p_0$ was in $S$ why would $\min\{d(p_0,p):p\in S\} = 0?$ Is it just because it is the minimum? How about if $p_0 \in S$ then what will $\max\{d(p_0,p):p\in S\}$ be?

max is not really the issue. The minimum of a set of numbers is the smallest number in the set. If you take the open interval (0,1) it doesn't have a minimum. It does have an infimum which is 0 but that's not in the set. So it doesn't have a minimum. So if you put S to be the open interval (0,1) and $p_0=0$, then the minimum does not exist. If S were closed why is the situation different?

 

[PeroK]

Homework Statement

Prove that if $S$ is a nonempty closed subset of $E^n$ and $p_0\in E^n$ then $\min\{d(p_0,p):p\in S\}$ exists.

Don't forget that {d(p_0,p):p\in S\} is a non-empty subset of IR bounded below (by 0).

 

[Lee33]

Thank you Dick and PeroK! I think I understand now.