8
$\begingroup$

I have the following definition:

In a metric space $(X,d)$ an element $x \in X$ is called isolated if $\{x\}\subset$ X is an open subset

But how can $\{x\}$ be an open subset? There has to exist an open ball with positive radius centered at $x$ and at the same time this open ball has to be a subset of $\{x\}$ but how can this be if there is only one element?

I'm trying to wrap my head around this, but I can't figure it out. It doesn't make sense for metrics on $\mathbb{R}^n$ since each open ball with some positive radius has to contain other members of $\mathbb{R}^n$.

The only thing I could think of was that we have some $x$ with 'nothing' around it and an open ball that contains only $x$ and 'nothing' (even though a positive radius doesn't make sense since there is nothing), so therefore the open ball is contained in $\{x\}$. But I'm not even sure we can define such a metric space, let alone define an open ball with positive radius containing only $x$ and 'nothing'.

$\endgroup$
  • $\begingroup$ as an example take the discrete metric $d$ and look at the ball $B_{1/2}(x)$ of radius $1/2$ centered at $x$. $\endgroup$ – user124910 Jan 2 '17 at 7:44
  • 1
    $\begingroup$ You can also do this : Take the plane $\mathbb{R}^2$ with the usual metric and delete the open set $B((0,0),1)-\{(0,0)\}$, the deleted neighbourhood of the origin with radius $1$. Now $(0,0)$ is isolated. $\endgroup$ – Levent Jan 2 '17 at 7:55
7
$\begingroup$

Let $X$ be any non-empty set, and define a function $d:X\times X\to\Bbb R$ as follows: for $x,y\in X$,

$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 1,&\text{if }x\ne y\;. \end{cases}$$

You can easily check that this function $d$ is a metric on $X$; it is commonly called the discrete metric on $X$. Now observe that for if $x\in X$ and $0<r\le 1$, then

$$B(x,r)=\{y\in X:d(x,y)<r\}=\{x\}\;:$$

the set $\{x\}$ is the open $r$-ball centred at $x$ provided that $0<r\le 1$.

For a less trivial example, consider the set

$$Y=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$

with the metric that it inherits from the usual metric on $\Bbb R$. You can check that for each $n\in\Bbb Z^+$ we have

$$B\left(\frac1n,r\right)=\left\{\frac1n\right\}$$

provided that $0<r\le\frac1{n(n+1)}$; this is because the point of $Y$ closest to $\frac1n$ is $\frac1{n+1}$, and the distance between them is

$$\frac1n-\frac1{n+1}=\frac1{n(n+1)}\;.$$

$\endgroup$
  • 1
    $\begingroup$ The second example, using the usual metric of $\mathbb{R}$ is great because it also explains the motivation behind the word "isolated" intuitively. I'd add that the point $0$ is not isolated in this space. $\endgroup$ – JiK Jan 2 '17 at 14:48
11
$\begingroup$

Suppose your metric space is $\mathbb{Z}$. Then you can take a ball around the element $4 \in \mathbb{Z}$ of radius $\frac{1}{2}$. The only element of your metric space in that ball is $4$, so the ball is just the set $\{4\}$, so $\{4\}$ is open.

$\endgroup$
  • 7
    $\begingroup$ For the sake of precision, $\mathbb{Z}$ together with the metric $d(x,y) = |y-x|$. $\endgroup$ – user14972 Jan 2 '17 at 9:42
6
$\begingroup$

Take, for instance, your set to be the integers, with the metric inherited from the metric on $\mathbb{R}^1$. Then any singleton is open, and so every point is isolated. This is because, as you said, these points have 'nothing' around them (if you look close enough).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.