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show/hide this revision's text 3 deleted 4 characters in body

Hello!

I recently started (it's purely self-education) reading a "Mathematical programming and optimizations" book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:

I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.

Assume we have some set $S \subset \mathbb{R_n}$ and for this set: $S = \overline{S}$ (closure set closure equals the set itself).

Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set $S :$

$\forall y \in \mathbb{R_n}, \space y \notin S: \space \exists ! \space p = \pi_S(y) $

This should mean that $S$ is a convex set.

Could someone please point me if I'm wrong (or right, but with some limitations for this statement) and help me proving it if I'm right.

I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)

Thank you.
show/hide this revision's text 2 pi_s -> pi_S

Hello!

I recently started (it's purely self-education) reading a "Mathematical programming and optimizations" book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:

I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.

Assume we have some set $S \subset \mathbb{R_n}$ and for this set: $S = \overline{S}$ (closure closure equals the set itself).

Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set $S :$

$\forall y \in \mathbb{R_n}, \space y \notin S: \space \exists ! \space p = \pi_s(y) pi_S(y) $

This should mean that $S$ is a convex set.

Could someone please point me if I'm wrong (or right, but with some limitations for this statement) and help me proving it if I'm right.

I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)

Thank you.
show/hide this revision's text 1

Convex sets and projections

Hello!

I recently started (it's purely self-education) reading a "Mathematical programming and optimizations" book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:

I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.

Assume we have some set $S \subset \mathbb{R_n}$ and for this set: $S = \overline{S}$ (closure closure equals the set itself).

Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set $S :$

$\forall y \in \mathbb{R_n}, \space y \notin S: \space \exists ! \space p = \pi_s(y) $

This should mean that $S$ is a convex set.

Could someone please point me if I'm wrong (or right, but with some limitations for this statement) and help me proving it if I'm right.

I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)

Thank you.