Perfect sets and topological vs. limit closure
Given a topological space $(X,\tau)$ and some subset $Y\subseteq X$, the subset $Y$ is perfect when it is closed and dense-in-itself.
Consider $X = \mathbb{Q}\cap[0,1]$ equipped with the topology imbued by the Euclidean metric, and the subset $Y = X$. $Y$ should be dense-in-itself, but cannot be perfect—as it is countable, and every perfect set is uncountable. It is also intuitive (but perhaps not true) that $Y$ doesn't satisfy the right "closure" properties under limits; that is, I should be able to construct a sequence of elements of $Y = \mathbb{Q}\cap[0,1]$ that converges onto an irrational number, and therefore not in $X$ or $Y$.
However, $Y$ is a closed set (in the topological sense) of $X$, as its complement is the empty set and the empty set is open by definition. Therefore, one (or more) of these two things is true:
- The topological notion of closedness doesn't agree with the notion of closedness with respect to taking limits in this case. This is what makes the most sense, but contradicts mathematical sources like this page.
- $\mathbb{Q}\cap[0,1]$ is not actually dense-in-itself. (I don't believe this to be true, although intuitively, there "shouldn't be enough points in $Y$" for every open set/neighborhood to contain more than one element due to its countability.)
Which is it? And if the first option is the issue, for what topological spaces does topological closedness imply closure with respect to limits and viceversa, if any?
Solution 1:
There is no contradiction at all: $Y$ is not perfect in $[0,1]$ (as it's indeed not closed, as can be shown by the sequence argument) but it is perfect in $Y$ (closed in $Y$, trivially, and still dense in itself).
"Perfect subset" is a relative notion: it depends on the space you're in. The dense in itself (i.e. having no isolated points, often called "crowded" in common parlance) is "intrinsic" and does not change on the surrounding space (if two subsets are homeomorphic (in their relative topologies), one is dense-in itself iff the other is. This does not hold for closed though: that's a relation vis-à-vis a larger space.
So you have to make clear what the context space is when talking about perfect subsets.
Historical side note
The traditional (going back to Cantor) way to define perfect sets in a space $X$ to define the set of limit points of $A$ (aka as derived set) by
$$A' = \{x \in X: \forall O \ni x \text{ open in } X: O \cap A\setminus\{x\}\neq \emptyset\}$$ and call a set perfect iff $A=A'$. The fact that $A \subseteq A'$ is a reformulation of dense-in-iself, no point of $A$ is isolated and $A' \subseteq A$ is equivalent to $A$ being closed in $X$. This way the seemingly arbitrary definition you stated becomes more natural, I think. Cantor introduced this notion because he was interested in iterated derived sets $A,A',A''=(A')',A''',\ldots$ for reasons related to convergence of triginometric series, and he became interested in sets where the process halts so when $A=A'$.
I hope this helps.