# Must a nonempty perfect set be uncountable?

Rudin's analysis book says that every "nonempty perfect set is uncountable".

But I'm confused -- consider the set $S$ of rational numbers in the interval $[1/3, 1/2]$.

It seems to me that $S$ is a nonempty perfect set (it is closed and all points are limit points), but it is countable.

What's wrong with this counterexample?

• What is your definition of closed? – Chickenmancer May 20 '17 at 17:13
• Once again someone has voted to close a newbie's question that is obviously about mathematics on the absurd grounds that it does not appear to be about mathematics. Maybe some people don't care if this group is made to appear to newcomers to be stupid or dishonest. As long as they don't have to sign their names to their opinions. – Michael Hardy May 21 '17 at 3:39

The theorem in Rudin says a nonempty perfect set in $\mathbb R^k$ is uncountable. A perfect set by definition is closed. "Closed" in the theorem stated in Rudin's book would have to mean "closed in $\mathbb R^k$". The set you've exhibited is closed in $\mathbb Q$ but not in $\mathbb R.$ No nonempty perfect set in $\mathbb Q^k$ is uncountable. So "perfect" has to mean "perfect in $\mathbb R^k$." Just as the property of being closed is relative to the ambient space, so also "perfection" is relative to the ambient space.

• thank you so much!! i really appreciate your help – J.Kim May 21 '17 at 10:30
• i would like to vote, but i can't..... – J.Kim May 21 '17 at 10:35
• @J.Kim : I'm glad it helped. – Michael Hardy May 22 '17 at 18:12

It's not closed. Pick any irrational in the interval (say, $0.4+{\pi\over 100}$) - there is a sequence of rationals approaching it.

• The set in question is indeed closed in $\mathbb Q.$ The resolution must lie in the fact that the theorem in Rudin's book says "in $\mathbb R^k$". – Michael Hardy May 20 '17 at 17:19
• @MichaelHardy Of course, but it's clear that the OP is working inside the reals; so I feel that introducing relative topologies will confuse more than help, in this case. – Noah Schweber May 20 '17 at 21:24
• But that was the very point at issue. – Michael Hardy May 21 '17 at 3:05
• @MichaelHardy Where do you get that? It looks like the OP just didn't see that $[{1\over 3}, {1\over 2}]\cap\mathbb{Q}$ is not closed in the usual sense of the word (that is, in $\mathbb{R}$ with the standard topology). I don't really see where they're switching between topological spaces, so I don't see why to bring that in here. – Noah Schweber May 21 '17 at 3:13
• It looked as if the poster thought being closed in $\mathbb Q$ and having no isolated points satisfied the hypotheses of the theorem. – Michael Hardy May 21 '17 at 3:25