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Let $f$ be a continuous function on $\beta\omega$ with $f(x_0)=0$ for some non-principal ultrafilter $x_0\in \beta\omega$. Is there an integer (a principal ultrafilter) $n$ such that $f(n)=0$?

EDIT: Since I am no longer permitted to give comments under my posts, I'd like to modify my question here:

Are there any first-countable compact Hausdorff spaces or at least compact Hausdorff spaces with a point $x_0$ having a countable nbhd basis such that any real continuous function from this space with $f(x_0)=0$ vanishes at some isolated point distinct with $x_0$?

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    @Batykaf: the system was not able to recognize the two accounts you used as one and the same because (a) you used two different e-mail addresses and (b) your accounts are **unregistered**. By registering your account the server can better keep track of question ownership and prevent instances where you cannot edit/comment on your own questions.2012-01-15

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NOTE: I assumed that the question was about real valued functions. (The OP did not specify this and this seemed to me as the most probable explanation.)


Not necessarily.

Note that if $f: \omega \to \mathbb R$ is any sequence convergent to $0$, then the continuous extension $\overline f:\beta\omega \to \mathbb R$ fulfills $\overline f(x)=0$ for every free ultrafilter (=for every point $x\in \beta\omega\setminus\omega$).

There are probably many ways how you can see that the above holds. (Depending on your favorite definition of $\beta\omega$, some of them might be clearer for you then others.) For instance, you can notice that $\overline f(x)$ is a cluster point of the sequence $(f(n))$ and in the case of convergent sequence, there is only one cluster point.

Thus choosing $f(n)=\frac1n$ gives an easy counterexample.

(Note that I am identifying integers with principal ultrafilters, which is quite usual in this context.)


About your edited question:

Are there any first-countable compact Hausdorff spaces or at least compact Hausdorff spaces with a point $x_0$ having a countable nbhd basis such that any real continuous function from this space with $f(x_0)=0$ vanishes at some isolated point distinct with $x_0$?

Every compact Hausdorff space is is normal. In a normal space, every closed $G_\delta$-set is zero-set, i.e. it is equal to $f^{-1}(0)$ for some real-valued function. This is a consequence of Urysohn's lemma. See e.g. Theorem 4 in Henno Brandsma's Useful theorems on normal spaces.

Now $\{x_0\}$ is a closed set and, since you also assume that it has a countable neighborhood basis, it is also a $G_\delta$-set. So there exists a real valued function on $X$ such that $f^{-1}(0)=\{x_0\}$.

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    @Brian Thanks for correcting my mistake, I've overlooked that the limit has to be 0. (But of course, that was what I meant.) About guessing codomain - I've worked with $\mathbb R$ and planned to edit my post if the OP clarifies that he meant something else.2012-01-13