I am a postgraduate student of physics. While doing some research on Poincare's work on the integrability of the three body problem, I came up with the following problem (which I feel unable to handle, possibly due to my insufficient background in general topology):

Let $X$ be a topological space and consider a non-constant, continuous function $f:X\to\mathbb{R}$ ( where $\mathbb{R}$ is considered with its usual euclidean topology). Is the inverse image of the rational values $f^{-1}(\mathbb{Q})\subseteq X$, always a dense subset of the domain $X$?

I would appreciate any help. Sorry in advance if this is not really research level.

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    $\begingroup$ Do you want to exclude examples like $f(x)=\pi$ and other constant valued irrational functions? These are trivially continuous and have $f^{-1}(\mathbb{Q})=\emptyset$ which isn't ever dense in $X$ unless $X=\emptyset$. $\endgroup$ – Alec Rhea Sep 20 '18 at 0:49
  • $\begingroup$ yes you are right. i mean non-constant functions. I edited that. $\endgroup$ – olgchar Sep 20 '18 at 0:53
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    $\begingroup$ Even if $f$ isn't globally constant, there could be an open subset of $X$ on which $f$ takes the value $\pi$ $\endgroup$ – Julian Rosen Sep 20 '18 at 1:19
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    $\begingroup$ Maybe if the data is analytic (e.g., $X$ an analytic manifold and $f$ a real analytic function) there's more hope? $\endgroup$ – Todd Trimble Sep 20 '18 at 1:26

Here is a refinement of your question that has a positive answer. Suppose $X$ has a dense subset $U$ that admits a differentiable manifold structure, and that there is a dense subset $V$ of $U$ on which $f$ is differentiable with non-vanishing derivative. Then $f^{-1}(\mathbb{Q})$ is dense in $U$, hence it is dense in $X$.


Not always. You need some assumptions on X (for example, X is connected) to make some conclusions about the inverse image. For example, if X is the irrationals and f is the inclusion map into the reals which looks like the identity, then your inverse image of the rationals is also empty.

Since you are studying dynamics possibly you want X to look like a real finite dimensional manifold, in which case the answer could be yes. Until I know more about X and f I can't really say.

Gerhard "Is Sometimes A Dense Set" Paseman, 2018.09.19.

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    $\begingroup$ It's going to take a lot more patching than that even for nice spaces $X$ like manifolds. If $f$ is constant irrational just on an open set (and extended however you please to the rest of $X$), the inverse image of the rationals will miss that open. $\endgroup$ – Todd Trimble Sep 20 '18 at 1:21
  • $\begingroup$ @Todd, indeed, the problem is under specified. I think the constant map case had been mentioned in the comments, and it seems the poster is interested in a subset of X which is dense in a large open subset of X, and not necessarily dense in all of X. Gerhard "Hopefully We Will Learn More" Paseman, 2018.09.19. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 1:35

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