# Every nonempty perfect set in $\mathbb R^k$ is uncountable: Rudin's argument

Here is Rudin's argument for the above theorem:

My question:

In the last paragraph Rudin says: "Since ${\mathbf x}_n \notin K_{n+1}$, no point of $P$ lies in $\bigcap_{1}^\infty K_n$". But why? $n$ goes to $\infty$ on both sides at the same pace, which implies that there still can be a ${\mathbf x}_n$ in the smallest $K_n$. May someone clarify this?

• There is no "smallest $K_n$". Jan 14, 2018 at 12:58

Suppose there is some point $p$ from $P$ in $\bigcap_{n=1}^\infty K_n$. As $P$ was enumerated as $x_1 ,x_2, x_3, \ldots$ (this is not a limit, it just expresses the fact that all points of $P$ occur as some $x_n$, it is just the expression of a bijection with $\mathbb{N}$), there is some $n$ (depending on $p$) such that $p= x_n$. But $p=x_n \notin K_{n+1}$ by construction. So $p=x_n \notin \cap_{n=1}^\infty K_n$. This last set is not a limit, it's just the set of all points that lie in all $K_n$ and we now found a concrete $K_n$ that $p$ does not lie in, so $p$ is by definition not in this intersection. So this contradiction shows that no $p\in P$ can lie in $\bigcap_{n=1}^{\infty} K_n$. But this in turn contradicts a theorem, and so the enumeration of $P$ cannot exist, and $P$ is uncountable. So it's a proof by contradiction inside a proof by contradiction. But perfectly valid.
As the $K_n$ are strictly decreasing there is no smallest $K_n$. E.g. $K_{100}$ is not the smallest, as $K_{101}$ is even smaller and so on. There also is no largest natural number,; we can always add $1$ to get a larger one.