# Perfect Set and Compact Set

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?

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.

• It may also be worth mentioning that every nonempty perfect set in $\mathbb{R}^n$ contains a compact perfect subset. Dec 30, 2012 at 5:14
• @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. Jan 2, 2013 at 21:20