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I came across this question and I don't have any idea where to start, so any help would be greatly appreciated!

The question is as follows: "Prove that $\mathbb{Q}$ with its usual topology can be embedded into a power of the two point discrete space".

I thought for a moment to look at it from a categorical product point of view but this got a little messy so is there something I am overlooking or an easier way to approach this?

Many thanks!

katherinebarry
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2 Answers2

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You have a countable family of maps $\mathcal C:=\{f_{n,q}:\mathbb Q\to\{0,1\}\mid n\in\mathbb N,q\in\mathbb Q\}$ where $f_{n,q}$ is the characteristic function of the open interval with radius $1/n$ centered at $q$. This family separates points and $\mathbb Q$ has the initial topology with respect to this family. The embedding theorem then implies that the evaluation map $e:\mathbb Q\to\{0,1\}^{\mathcal C}$ which is defined $e(x)(f)=f(x)$ is an embedding.

Stefan Hamcke
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  • so helpful, thank you! – katherinebarry Apr 11 '13 at 21:56
  • Don't you need separation of points and closed sets, not just points? – Henno Brandsma Apr 12 '13 at 09:09
  • @HennoBrandsma: No separation of points is enough. Well, a Tychonoff space $X$ can be embedded in a cube, but the separation of points and closed sets in that case gives the space the initial topology. Then you have separation of points because points are closed. These two conditions are sufficient for the embedding. – Stefan Hamcke Apr 12 '13 at 10:20
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    @StefanH.: Could you please state or give a reference for those embedding theorem? – Dune Apr 12 '13 at 10:44
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    @Dune. The statement is: The evaluation map $e:X\to\prod_{\alpha\in A}X_\alpha$ is an embedding if and only if the collection ${f_\alpha:X\to X_\alpha\mid\alpha\in A}$ separates points and $X$ has the initial topology with respect to this collection. It can be found for example in Willard's General Topology in chapter 8. – Stefan Hamcke Apr 12 '13 at 12:49
  • @StefanH.: Great, thank you! – Dune Apr 12 '13 at 12:55
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I will give an embedding of $\mathbb Q\cap (0,1)$ into $2^{\mathbb N}$.

$\mathbb Q $ is homeomorphic to $\mathbb Q\cap (0,1)$, so it is enough to embed the latter set.

Let $\alpha$ be irrational. Map each $q\in \mathbb Q\cap [0,1]$ to the binary representation of $q+\alpha$ mod 1, i.e., to the binary representation of the fractional part of $q+\alpha$.

(Note that the natural map from $2^{\mathbb N}\setminus A$ onto $Q\cap [0,1]\setminus B$ is a homeomorphism if you let $A$ and $B$ be the "dyadic rationals".)

g.castro
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