# Clarification: proof that perfect subsets of R^k uncountable

1. Feb 12, 2016

### cpsinkule

From Baby Rudin
"Thm: Let P be a non-empty, perfect subset of R^k. Then P is uncountable.

Pf: Since P has limit points, P must be infinite. Suppose P is countable, list the point of P {x1 ...xn }. Construct a sequence of nbhds. as follows. Let V1 be any nbhd of x1 . Suppose Vn has been constructed so that Vn ∩ P is non-empty. Since every point of P is a limit point of P, there is a nbhd Vn+1 such that it's closure is in Vn , xn is NOT in the closure of Vn+1, and Vn+1 ∩ P is non-empty. Let Kn=closure(Vn)∩P, then Kn is closed, bndd., therefore compact. The Kn are clearly nested compact sets, thus their arbitrary intersection is non-empty. But xn∉Kn+1 implies that ∩nKn is empty, thus a contradiction. "

My confusion is in the claim of the contradiction. How does xn∉Kn+1 imply the intersection is empty? My thoughts are that, since we assumed P to be countable, xn must be an element of any inductive subset of P indexed by n+1. Is my logic behind this statement true? It just seemed a little artificial to me as I have not seen a proof which uses induction to arrive at a contradiction in this manner before.

2. Feb 12, 2016

### Staff: Mentor

What do you mean by this? $X = \{x\}$ has limit points, $\{x\}$.
What does it mean for P to be perfect? Beside there is a point in P and P has nice topological separation properties that is all you have.

3. Feb 12, 2016

### Krylov

This is not true.
A set is perfect when it is closed and each of its point is a limit point.

4. Feb 12, 2016

### Staff: Mentor

I've read closed and no isolated points.
Why is a constant sequence not allowed? Likely I've mistranslated limit point by limit. It's kind of a standard pit.

5. Feb 12, 2016

### Krylov

Because $x$ is a limit point of a set $A$ in a metric space if, by definition, every neighborhood of $x$ meets $A$ in a point other than $x$.

All limit points of $A$ are in $\overline{A}$, but not all points in $\overline{A}$ are necessarily limit points of $A$.

Last edited: Feb 12, 2016
6. Feb 12, 2016

### Hawkeye18

If you denote $K=\cap_{n\ge 1} K_n$, then the condition $x_1\notin K_2$ implies that $x_1\notin K$. Similarly, condition $x_2\notin K_3$ implies that $x_2\notin K$, and, more generally, for every $n\ge 1$ the condition $x_n\notin K_{n+1}$ implies that $x_n\notin K$. But this means that $P\cap K=\varnothing$ (because $P=\cup_{n\ge 1} \{x_n\}$). On the other hand, by construction $K\subset P$ (because $K_n\subset P$ for all $n\ge 1$).

So $K=\varnothing$.

7. Feb 12, 2016

### cpsinkule

Thanks, that's much clearer now! :)