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Let $f\colon \beta \mathbb{N}\to \mathbb{C}$ be a continuous function such that $f(x)=0$ for some $x\in \beta\mathbb{N}\setminus \mathbb{N}$. Can we conclude that there exists both closed and open subset $D\subseteq \beta\mathbb{N}$ such that $f(y)=0$ for all $y\in D$? Can $D$ be homeomorphic to $\beta \mathbb{N}$?

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Of course it can be homeomorphic, for example $f(x)=0$ for all $x$... – Asaf Karagila Oct 1 '12 at 15:38
I mean: if such a $D$ always exists, can it be taken to be homeomorphic to $\beta \mathbb{N}$. – zirko Oct 1 '12 at 15:40
up vote 2 down vote accepted

The answer is yes on all counts.

Let $0=f(p)$, where $p\in\beta\Bbb N\setminus\Bbb N$, and for $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:|f(k)|<2^{-n}\}$; then $U_n\in p$ for each $n\in\Bbb N$. Clearly $f^{-1}[\{0\}]=\bigcap_{n\in\Bbb N}\widehat{U_n}$, where for any $A\subseteq\Bbb N$, $\widehat A=\{q\in\beta\Bbb N:A\in Q\}$.

Recursively construct distinct $n_k\in\Bbb N$ so that $n_k\in U_k$ for each $k\in\Bbb N$, and let $U=\{n_k:k\in\Bbb N\}$; clearly $U\setminus U_n$ is finite for each $n\in\Bbb N$, so for any $q\in\beta\Bbb N\setminus\Bbb N$, $U\in q$ iff $U\cap U_n\in q$. Suppose that $q\in\widehat U$; then $U_n\supseteq U\cap U_n\in q$ for each $n\in\Bbb N$, so $q\in\bigcap_{n\in\Bbb N}\widehat{U_n}$, and therefore $f(q)=0$. Thus, $f[\widehat U]=\{0\}$.

Finally, $\widehat U$ is a clopen subset of $\beta\Bbb N$ and is homeomorphic to $\beta\Bbb N$.

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Thank you very much. – zirko Oct 1 '12 at 22:28
@zirko: My pleasure. – Brian M. Scott Oct 1 '12 at 22:32

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