0
$\begingroup$

Let $(X,d)$ be a metric space.

Given $p ∈ X$ and $E ⊂ X$ we define the distance from $p$ to $E$ as the number

$$d(p,E) = \inf\{d(p, x) : x ∈ E\}.$$

Show that if $p$ is an accumulation point of $E$ then $d(p,E) = 0$.

I have a sense that it should be 0 because we take inf of distance, but I have no idea for how to show it.

  • 0
    E is closed or open?. In any case $p \in \overline{E}$. – 2017-01-25
  • 0
    @manhattan : doesn't matter. True for any $E$. – 2017-01-25

2 Answers 2

0

It is known that $\overline{E}=\{x\in X:d(x,E)= 0\}$(See for instance Chapter 8 of Schaums Outline Series in General Topology).

Now, if $p$ is an accumulation point of $E$, then certainly, $p\in\overline{E}$ and so $d(p,E)=0$.

0

Hint: Suppose $p$ is an accumulation point of $E$. Do you see that this means there are points of $E$ distinct from $p$ which are arbitrarily close to $p$? And what is the infimum of a set of positive numbers which contains arbitrarily small positive numbers?