I can't figure out this paragraph:
$E$ is open relative to $Y$ if to each $p\in E$ there is associated an $r\gt 0$ such that $q\in E$ whenever $d(p,q)\lt r$ and $q\in Y$
Does this look like this:
$\forall p\in E, \exists r\in\Bbb R^+$ such that $d(p,q)\lt r, q\in Y \implies q\in E$
I can't figure out what this means, can I have some examples?
My attempt: For some $E$ open relative to $Y$, $E$ is bounded by some $r$, e.g. $\forall p,q\in E,\exists r\in\Bbb R^+,d(p,q)\lt r$ and every one of these elements is also in $Y$ or something.
I can't work it out, someone explain please?
A set $A$ is open relative to $E$ if given any $x\in A$ there is a ball $B(x,\delta)$ such that $B(x,\delta)\cap E $ is contained in $A$. This simply means that we don't want the whole ball to be contained in $A$; but just the part of the ball that is inside $E$. For example, $[0,1)$ is open relative to $[0,1]$, but it is not open relative to $\Bbb R$.