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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: You should edit your question and add the information about the codomain. (So that other users don't need to look into comments.) Notice that my answer only covers the situation $f:\beta\omega\to\mathbb R$, so if you meant something else, you should not have accepted that answer. –  Martin Sleziak Jan 13 '12 at 10:51
    
As the OP writes I am no longer permitted to give comments under my posts, the two accounts math.stackexchange.com/users/22890/batykaf and math.stackexchange.com/users/22878/batykaf should probably be merged. I am flagging the question for moderator's attention. –  Martin Sleziak Jan 14 '12 at 5:54
    
@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. –  Willie Wong Jan 15 '12 at 18:31
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1 Answer 1

up vote 3 down vote accepted

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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Thanks. Are there any first-countable compact spaces or at least compact spaces with a point $x_0$ with a countable nbhd basis such that any real continuous function from this space with $f(x_0)=0$ vainshes at some isolated point distinct with $x_0$? –  Batykaf Jan 13 '12 at 10:31
    
You want $f$ to be a sequence converging to $0$, not just any convergent sequence. (How did you guess that $\mathbb{R}$ and not $\beta\omega$ was the intended codomain?!) –  Brian M. Scott Jan 13 '12 at 10:33
    
@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. –  Martin Sleziak Jan 13 '12 at 10:39
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