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"Since $\mathbf{x}_n \notin K_{n+1}$, no point of $P$ lies in $\bigcap_{1}^\infty K_n$".

As far as I understand, $K_n$ is a compact subset of a perfect set $P$ which is assumed as a countable set, $P = \{\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3,\dots\}$, and $K_{n+1}$ is constructed such that $\mathbf{x}_n \not\in K_{n+1}$ and $K_{n+1} \subset K_n$. The proof ends by showing the contradiction that $\bigcap_1^\infty K_n$ is both empty and nonempty.

I have a problem in understanding the sentence in the quotation marks. If we assume $P = \{1,1/2,1/3,\dots\}\cup\{0\}$ and define $K_n = [1/n,0]$, then clearly $\bigcap_{1}^\infty K_n = \{0\}$. What am I missing here?

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  • $\begingroup$ The issue is that in defining $P$, you enumerated the elements of $P$. For your example, $0$ must fall in this enumeration somewhere; you can't list $0$ after listing $1/n$ for all $n$. This invalidates your counterexample, because if $0$ is now listed as the $n$-th element, it cannot lie in $K_{n+1}$. $\endgroup$
    – Andrew
    Jun 6, 2016 at 6:26

1 Answer 1

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Since $x_n\notin K_{n+1}$, it is not in the intersection of all the $K_n$. Thus $P$, which is made of points $x_n$, has no points in the intersection.

Your example is not a counterexample because you have not indexed $P$ with $1,2,3,\ldots$. If you insert the $x_N=0$ somewhere in the sequence $1,1/2,1/3,\ldots$, then suddenly for all $n\geq N$, you have $x_n\in K_{n+1}$.

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  • $\begingroup$ Thanks. The problem has been resolved. $\endgroup$
    – flyingwith
    Jun 6, 2016 at 6:33

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