I am having some difficulty in understanding the difference between Perfect and Compact sets. More specifically, my problem is rather understanding how Perfect sets are different from Compact sets, by that I mean, I understand Compact Sets more than Perfect Sets.

I know that for any set, $S$, to be Compact, every sequence of $S$ has a subsequence that converges to a point which also lies in $S$. This is basic definition but is not difference than saying it is Bounded and Closed or Heine Borel Theorem.

Now, the definition of Perfect Set is $P$ is a Perfect Set if $P =P'$ where $P'$ is the set of Limit Points of $P$ (WolframAlpha). At other places, I also that a set is Perfect Set if $P$ is closed and accumulation point of $P$. Though, I do not understand this completely, it sounds similar to definition of Compact Sets.

I would appreciate any explanation.

Further, are there are any non-singleton sets that are Compact but Not Perfect? How about Perfect but Non Compact Sets?


1 Answer 1


What you give as the definition of compact set is actually the definition of sequentially compact set; the two properties are equivalent in metric spaces but not in general. In general a set $K$ in a topological space $X$ is compact if every open cover of $K$ has a finite subcover. This means that if $\mathscr{U}$ is a family of open sets such that $K\subseteq\bigcup\mathscr{U}$ (i.e., $\mathscr{U}$ is an open cover of $K$), then there is a finite subcollection $\mathscr{U}_0\subseteq\mathscr{U}$ such that $K\subseteq\bigcup\mathscr{U}_0$. This no longer looks at all like the definition of a perfect set.

Your definition of perfect set is correct: $P$ is perfect if $P=P'$. In Hausdorff spaces (and hence certainly in metric spaces) this is equivalent to saying that $P$ is an infinite closed set with no isolated points.

Even in metric spaces the two are definitely not the same. This can already be seen in the familiar space $\Bbb R$. The set

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

for example, is an infinite compact subset of $\Bbb R$ that is certainly not perfect: $S$ consists mostly of isolated points, and $S'=\{0\}$. On the other hand, $\Bbb R$ itself is a perfect subset of $\Bbb R$ that is not compact. The closed ray $[0,\to)$ is another, as is

$$\bigcup_{n\in\Bbb Z}[2n,2n+1]=\ldots\cup[-4,-3]\cup[-2,-1]\cup[0,1]\cup[2,3]\cup[4,5]\cup\ldots\,\;.$$

What is true in $\Bbb R$ (and in fact in $\Bbb R^n$ for all $n$) is that every uncountable closed set contains a perfect subset.

  • $\begingroup$ It may also be worth mentioning that every nonempty perfect set in $\mathbb{R}^n$ contains a compact perfect subset. $\endgroup$ Dec 30, 2012 at 5:14
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    $\begingroup$ @Trevor Wilson: In fact, from every nonempty perfect set in ${\mathbb R}^n$ you can select continuum many (i.e. $2^{\aleph_{0}}$ many) compact perfect sets that together form a pairwise disjoint collection of sets. Note the contrast with the situation when "perfect set" is replaced with "$n$-dimensional interval", or with "open set", when continuum many becomes at most countably many. $\endgroup$ Jan 2, 2013 at 21:20

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