For a point p ∈ Rd and a set S ⊂ Rd, we define the distance from p to S to be.. Show that p is a limit point of S if and only if ρ(p,S) = 0.

Question

For a point $p ∈ R^d$ and a set $ S ⊂ R^d$, we define the distance from p to S to be

$p(p, S)$ = inf$ ∥p − q∥, q∈S$

By convention, we set p(p,∅) = ∞.

Show that p is a limit point of S if and only if ρ(p,S) = 0.

I'm very confused. Why would that only be true if it equals 0? Can someone explain this question to me? Topology in general is such a bad unit for me.

Answer 1

If $p$ is a limit point of $s$ then there exists a sequence $(s_n)$ in $S$ such that $d(p,s_n) \to 0$. Since $0 \leq d(p,S)\leq d(p,s_n)$ for all $n$ we get $d(p,S)=0$ by taking limit as $ n \to \infty$.

Conversely, $d(p,S)=0$ implies (by definition of infimum of a set of real numbers) that there exists a sequence $(s_n)$ in $S$ such that $d(p,s_n) \to 0$. This implies that $p$ is a limit point.

Answer 2

The statement is trivially true for $S=\emptyset$. $\emptyset$ has no limit points and no points exist with $p(p,S)=0$.

So assume $S\neq \emptyset$ and let $p$ be a limit point of $S$. It's clear that $p(p,S)\ge 0$ as $0$ is a lower bound for $\{\|p-q\|: q \in S\}$ and $\inf$ is the largest lower bound. Let $r>0$ be arbitrary. Then there is some $q_0 \in S$ such that $\|p-q_0\|< r$ (as $B(p,r)$ intersects $S\setminus \{p\}$) and so $p(p,S) < r$ (as it is a lower bound for $\{\|p-q\|: q \in S\}$ and so $$p(p,S)=\inf \{\|p-q\|: q \in S\} \le \|p-q_0\| < r$$

So $p(p,S) =0$ as is any real number $w\ge 0$ that is smaller than any $r>0$.

On the other hand if $p(p,S)=0$, $p$ is an adherent point of $S$: if $r>0$ we must have that $B(p,r) \cap S \neq\emptyset$ because if this is not the case we know

$$\forall q \in S: \|q-p\|\ge r$$ and thus

$$p(p,S)=\inf \{\|p-q\|: q \in S\} \ge r>0$$ ($r$ is a lower bound and inf is the largest lower bound), contradicting $p(p,S)=0$.

We cannot conclude it is a limit point, check $p=2$ and $S=[0,1]\cup \{2\}\subseteq \Bbb R$.