Definition of Perfect Set says

$E$ is perfect if $E$ is closed and if every point of $E$ is limit point of $E$.

Is it possible that every point of $E$ is limit point of $E$ but $E$ is not closed ? I'am unable to find an example. Because Closed set is defined as if every limit point of $E$ is a point of $E$.

If not possible. Then what is need of including in Definition of Perfect set ? ( that If $E$ is closed ) Help would be appreciated !

up vote 11 down vote accepted

Yes, of course. Note that $\Bbb Q$ is not closed in $\Bbb R$, but every rational number is the limit of a sequence of rational numbers, all different than your proposed limit.

In other words, $\Bbb Q$ has no isolated points, but it's not closed as a subset of $\Bbb R$.

take $(a,b)\subset\mathbb{R}$. If $x\in(a,b)$ then $\exists a_n\in(a,b):a_n\to x$

  • 3
    $x\in(a,b)$ then $\exists a_n\in(a,b):a_n\to x$ isn't enough for $x$ to be a limit point. – JiK Aug 14 at 17:15

Others have already given nice examples.

About the notion: $E$ is closed if $E$ contains its limit points, or equivalently $E' \subseteq E$.

A set is perfect when moreover $E \subseteq E'$ (all points are limit points, or there are no isolated points) so iff $E' = E$. So it's a fixed point of the derived set operator $E \to E'$, so to speak. This indeed implies closedness by the above criterion for closedness.

This notion is natural to consider when we start with some closed subset $E$ and form the succesive derived sets $E, E', E'', \ldots$ or more formally define $E^{(0)} =E$, $E^{(\alpha+1)} = (E^{(\alpha)})'$ for all succesor ordinals $\alpha+1$, and $E^{(\alpha)} = \bigcap_{\beta < \alpha} E^{(\beta)}$ when $\alpha$ is a limit ordinal. One can show that for some ordinal $\gamma$, $E^{(\gamma)} = E^{(\gamma+1)}=: K$, and this $K$ is called the perfect kernel of $E$. Such considerations led Cantor to define ordinals and to consider perfect subsets of the reals.

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