Do all rationals $\left ( \mathbb{Q} \right )$ form a perfect set if the metric space is $\mathbb{Q}$?

Question

The following question came to my mind while reading the definition of a perfect set from Walter Rudin's Principles of Mathematical Analysis, 3rd Edition, Page 32, Section 2.18(h):

If the metric space is the set of all rational numbers $\mathbb Q$ (instead of $\mathbb R^1$ -- the set of all real numbers), then does $\mathbb Q$ constitute a perfect set?

In my understanding, the answer is "yes". Because $\mathbb Q$ is closed (since $\mathbb Q$ contains each of its limit points if the underlying metric space is $\mathbb Q$) and because every point of $\mathbb Q$ is a limit point of $\mathbb Q$ (arbitrary proximity). I'm not feeling so much sure somehow, and didn't find a corroboration elsewhere. Therefore this query.

Would appreciate help. Thanks.

Answer 1

Yes, that's exactly right.

Of course, most applications of perfect sets (that I'm aware of anyways) require the ambient space to be "reasonable," usually completely metrizable, and in such spaces perfect sets are very unlike $\mathbb{Q}$ (e.g. uncountable). But if we do take $\mathbb{Q}$ to be our ambient space, then yes, in that context $\mathbb{Q}$ is perfect.

(Note incidentally that perfectness makes sense, not just in the context of metric spaces, but more general topological spaces.)