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Motivation: My textbook states that a connected subset of a normal space is mapped to all of $[0,1]$ if it has non-empty intersections with disjoint closed sets, one of which is mapped to $\{0\}$ and the other mapped to $\{1\}$. This is because the image of a connected set has to be a connected set, by Darboux's lemma.

But on reading the proof to Urysohn's lemma, on which this assertion is based, I was under the impression that points were mapped only to rational points on $[0,1]$.

So isn't it true that the connected subset should have been mapped to $\Bbb{Q}\cap [0,1]$ rather than all of $[0,1]$?

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The proof of Uryson’s lemma doesn’t actually tell you exactly which points of $[0,1]$ are in the range of the function, except that $0$ and $1$ definitely are. At one extreme the range may be $\{0,1\}$, at the other it may be $[0,1]$, and it may be somewhere in between. If you look closely, you’ll see that while open sets $U_r$ are defined only for rational $r$, the values of the function itself are not necessarily labels on those sets $U_r$.

$\Bbb Q\cap[0,1]$ is clearly not connected, since it’s the disjoint union of the relatively open sets $\Bbb Q\cap\left[0,\frac{\sqrt2}2\right)$ and $\Bbb Q\cap\left(\frac{\sqrt2}2,1\right]$.

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  • $\begingroup$ True! What a mess. In fact I do remember thinking about the fact that even irrational numbers like $\sqrt{2}$ can be mapped to if that is the $\inf$ of a sequence of rational numbers converging to it. $\endgroup$ – fierydemon Jun 29 '13 at 2:23
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There must be some limiting process involved, since $\mathbb Q\cap [0,1]$ is not connected as $$\mathbb Q\cap [0,1]=(\mathbb Q\cap [0,\sqrt2/2))\cup (\mathbb Q\cap (\sqrt2/2,1])$$ and the sets on the RHS are open, disjoint and nonempty.

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