Question

Nearest point theorem: letF be a non-void closed subset of R^p and let x be a point outside of F. Then there exists at least one point y belonging to F such that ||z - x|| greater than or equal to ||y -x|| for all z in F.

Given the theorem, answer the question:

Does the nearest point theorem in R imply that there is a strictly positive real number nearest zero?

Answer 1

Let
d = inf{||2 – 3|| : IEF
​​​​​​.Since F is nonempty, by the definition of infimum for all
neN
, there exists
In EF
, such that
+ P> ||42 - 3|| SP
. Hence we can find a sequence
{n} CF
, such that
+ P> ||42 - 3|| SP
. Thus we have for all
neN
,  
1|2n|| = || In - I+x|| < ||2 - In|| + ||0|| < d + ||0|| + = <d+ ||2|| +1
. Hence we have a bounded sequence
{n} CF
. Then by Bolzano–Weierstrass theorem every bounded infinite set has a limit point, say , such that
d= 11 - y
. Note that y is also a limit point of F, by the construction of $\{x_n\}$. Since F is closed we have
Y EF
.

Now we have $d=||x-y||=\inf_{z\in F}\{||z-x||:x\in F\}\le ||z-x||$ , for all
ZEF
.

No the above theorem does not imply there exists strictly positive real number nearest to 0, as if possible there exists such , such that p is nearest to 0, then consider
<
, which contradicts that p is the nearest to 0.

Feel free to comment if you have nay doubts. Cheers!