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I am looking for a topology such that R is dense in C . I was thinking I can construct a surjective continuous function f : R → C such that the image of Q is R.

Marc van Leeuwen
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Ißra Jörg
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    $\mathbf{Q}$ is too small a set to be mapped surjectively onto $\mathbf{R}$. But, otherwise I think you can make use of the idea. I would drop the assumption that $f$ is continuous, a random bijection is likely to work. – Jyrki Lahtonen May 06 '17 at 05:47
  • I'm confused by the title? – marshal craft May 06 '17 at 06:01
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    Other than your method, finite-complement topology on $\Bbb C$ would also make $\Bbb R$ dense in $\Bbb C$. – Error 404 May 06 '17 at 07:46

3 Answers3

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There are many many possibilities; the following is probably the simplest. Take the indiscrete topology on $\mathbb{C}$, in which the only open sets are $\mathbb{C}$ and $\emptyset$. Then every nonempty subset of $\mathbb{C}$ is dense.

Eric Wofsey
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Take an automorphism of $\Bbb C$ other than the identity or complex conjugation. These exist by Zorn's lemma. The image of $\Bbb R$ under this automorphism is an isomorphic copy of $\Bbb R$ dense in $\Bbb C$.

Angina Seng
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    This works! The image contains two numbers linearly independent over $\Bbb{R}$, and the $\Bbb{Q}$-span of those two elements is A) contained in the image, B) dense in $\Bbb{C}$. – Jyrki Lahtonen May 06 '17 at 05:52
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    I agree +1, though I find this rather (a bit too) succinct. To answer the actual question (which is about $\Bbb R$ itself), you should say that the topology on $\Bbb C$ is taken to be the one where the open sets are the images by the inverse of the chosen automorphism of the usual open sets. It might also be helpful to indicate just how Zorn's lemma is used; this sounds a bit like "it is known to be true supposing the axiom of choice, so no doubt one can arrange to use Zorn's lemma somewhere on the way". – Marc van Leeuwen May 06 '17 at 09:45
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The Zariski topology is an important “real-world” topology having this property. In a complex affine space (maybe not just a complex line), the Zarisiki topology is the topology whose closed sets are the zeroes of polynomial functions. In dimension 1 the closed sets in the Zariski topology are just the finite subsets, together with the full complex line, which indeed makes the real line a dense subset.

The Unitarian trick is a great example of an argument involving distinct topologies on some space (in this case the Zariski topology and the Euclidean topology) to prove a meaningful theorem.

There is a lot of other topologies making the real line a dense subset of the complex plane, but only a few of them are interesting. Some already quoted the coarse topology. If you are interested in producing a lot of rather artificial examples, take any map $f$ from the complex plane onto a finite set $F$, so that the restriction of $f$ to the real line is also onto. The set of preimages through $f$ of the poert set of $F$ are the closed sets of a topology on the complex plane where the real line is dense.

Michaël Le Barbier
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