Is there a Hausdorff topological space $X$ such that for any continuous map $f: X\longrightarrow \mathbb{R}$ and any $x\in \mathbb{R}$, the set $f^{-1}(x)$ is either empty or infinite?
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1$\begingroup$ What happened to your other question, concerning spaces $X$ for which there is a function $f:X\to\mathbb{R}$ that is discontinuous at every point? I can't seem to find it now. $\endgroup$– Joel David HamkinsCommented Aug 9, 2016 at 23:14
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$\begingroup$ @JoelDavidHamkins I suppose it is this question: mathoverflow.net/questions/247146/… (I do not have sufficient reputation to see deleted posts. But at the moment the question can be still seen in Google cache.) $\endgroup$– Martin SleziakCommented Aug 13, 2016 at 10:15
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3 Answers
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Yes, there is such a space. Let $X=2^{\omega_1}$ be the space of binary sequences of length $\omega_1$, in the order topology generated by the lexical order. So $X$ consists of the branches through the tree $2^{<\omega_1}$, with the left-to-right order on branches. This is an order topology of a linear order and hence Hausdorff.
The key thing to notice is that every element $a\in X$ is the limit of an $\omega_1$-sequence. If $a_\alpha\to a$ for $\alpha<\omega_1$ and $f:X\to\mathbb{R}$ is continuous, it follows that $f(a_\alpha)\to f(a)$. Since every convergent $\omega_1$-sequence in the reals is eventually constant, it must be that $f(a_\alpha)=f(a)$ for all sufficiently large countable ordinals $\alpha$. So this space has your desired property.
In fact, I claim that for every continuous function $f:X\to\mathbb{R}$ and every $a\in X$, the function is constant on an interval about $a$. To see this, we may find open intervals in $\mathbb{R}$ so that $\{f(a)\}=\bigcap_n I_n$, and then $f$ is constant on $\bigcap_n f^{-1}I_n$. Each such $f^{-1}I_n$ is an open set containing $a$, and since there are only countably many, we may find an interval containing $a$ inside all of them.
Thus, this space has the property that every continuous function $f:X\to\mathbb{R}$ is locally constant, and every nonempty open set has size $2^{\omega_1}$.
answered Aug 9, 2016 at 12:10
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$\begingroup$ One could argue similarly with the surreal line (or the surreal line at some birthday of uncountable cofinality, to make it a set), which also has the property that every continuous function to the reals is locally constant, for essentially similar reasons. $\endgroup$ Commented Aug 9, 2016 at 12:55
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1$\begingroup$ For the record, the completely regular spaces $X$ where every function $f:X\rightarrow \mathbb{R}$ are locally constant are known as $P$-spaces. Equivalently, a $P$-space is a regular space where the countable intersection of open sets is open. $\endgroup$ Commented Aug 16, 2016 at 19:15
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How about this example: $[0,1]^A$ with the product topology, where $A$ is uncountable. Then every nonempty $G_\delta$ set is uncountable. Since your sets $f^{-1}(x)$ are $G_\delta$ sets, this has your property.
answered Aug 9, 2016 at 13:42
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$\begingroup$ Nice example, Gerald. It is similar to my example, but do you get the locally-constant property? It seems not, since the projection functions to the unit interval are continuous, but not locally constant. But if you used $2^A$ instead of $[0,1]^A$, then at least the projection functions would be locally constant. Is every continuous function locally constant in this case? $\endgroup$ Commented Aug 9, 2016 at 17:18
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$\begingroup$ Locally constant in $\{0,1\}^A$? No. Think of the projection onto $\{0,1\}^B$, where $B \subset A$ is countable. (In fact, the general continuous real-valued function factors through such a projection.) But we do have each nonempty level set $f^{-1}(x)$ has the same power as the whole space. $\endgroup$ Commented Aug 9, 2016 at 17:38
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If $(X,\mathcal{S})$ is a regular space, then the $P$-space coreflection of $(X,\mathcal{S})$ is the space generated by the basis $\{\bigcap_{n\in\omega}U_{n}\mid\forall n\,U_{n}\in\mathcal{S}\}$. A completely regular space $X$ satisfies the property that $f^{-1}(\{x\})$ is empty or infinite for any real $x$ if and only if the $P$-space coreflection of $X$ has no isolated points.
There exists infinite regular spaces $X$ where every continuous function $f:X\rightarrow\mathbb{R}$ is constant and in such spaces $f^{-1}(x)$ is always empty or infinite.
