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I am trying to find an example of a separable Hausdorff space which has a non-separable subspace. This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or non-separable?

A space is separable if it contains a countable dense subset. A subset $A$ of a space $X$ is dense in $X$ if $\bar{A}=X$.

It's easy to come up with a dense set and a countable set in $\mathbb{R}\setminus\mathbb{Q}$, since (trivially) $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}$ in the subspace topology, and as a countable set we can take something like $\{k\pi \, | \, k \in \mathbb{Z}\}$. But is there a subset that is both dense AND countable? And of course, how do we prove the result?

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For your original question, you can find a separable space with an uncountable discrete subspace. –  azarel Aug 14 '12 at 4:02
    
@azarel: But is it Hausdorff? –  Asaf Karagila Aug 14 '12 at 4:03
    
@AsafKaragila Sure, even Tychonoff. –  azarel Aug 14 '12 at 4:04
    
@azarel: Then this is not what I had in mind. Is your space compact? :-) –  Asaf Karagila Aug 14 '12 at 4:05
    
@AsafKaragila No, it is not compact but I guess you could get one. –  azarel Aug 14 '12 at 4:26
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2 Answers

up vote 12 down vote accepted

$\Bbb R$ is second countable (i.e., has a countable base), so it’s hereditarily separable. Specifically, let $\mathscr{B}$ be the set of all open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable. Enumerate $\mathscr{B}=\{B_n:n\in\Bbb N\}$, and for $n\in\Bbb N$ let $x_n$ be any irrational number in $B_n$. Then $\{x_n:n\in\Bbb N\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$. (Clearly the same trick works for any subset of $\Bbb R$, not just the irrationals.)

For an explicit example of such a set, let $\alpha$ be any irrational; then $\{p+\alpha:p\in\Bbb Q\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$.

There are many separable Hausdorff spaces with non-separable subspaces. Two are mentioned in this answer to an earlier question. The first is compact; the second is Tikhonov and pseudocompact. Both are therefore quite nice spaces. Both are a bit complicated, however. A simpler example is the Sorgenfrey plane $\Bbb S$: $\Bbb Q\times\Bbb Q$ is a countable dense subset of $\Bbb S$, and $\{\langle x,-x\rangle:x\in\Bbb R\}$ is an uncountable discrete subset of $\Bbb S$ (which is obviously not separable as a subspace of $\Bbb S$).

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Very nice, thank you. –  Alex Petzke Aug 14 '12 at 12:34
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Every subspace of a separable metric space is separable.

For the irrationals, take the irrational algebraic numbers, those are dense in $\mathbb R$ and therefore in the irrationals too. As Jacob remarks below, if $\alpha$ is irrational, then $\{\alpha+q\mid q\in\mathbb Q\}$ is also dense.

Generally speaking, the irrationals are homeomorphic to the Baire space, the set of sequences of natural numbers, equipped with the product topology $\mathbb N^\mathbb N$.

This has a countable dense subset: eventually constant sequences.

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Perhaps a simpler example of a countable dense subset of irrationals would be $\mathbb Q + \pi$. –  JSchlather Aug 14 '12 at 3:37
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@JacobSchlather: I have to say the first example I thought of was the algebraic numbers as well. –  Kris Aug 14 '12 at 3:39
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I was going to go with $\Bbb Q+\sqrt{2}$, personally. –  Cameron Buie Aug 14 '12 at 3:46
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@Cameron: This is true assuming countable choice. Separable metric spaces are second countable; if you take a subspace it is also second countable (relativize the base to the subspace) now choose a point from each basic open set and you have your precious separability! Without countable choice, take Cohen's first model and the D-finite subset of $\mathbb R$ as a counterexample. –  Asaf Karagila Aug 14 '12 at 3:56
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@CameronBuie. To add on Asaf's comment, it's worth noting that the word 'metric space' plays a key role there. Metrizability is hereditary and so is second countability. Since in a metric space separability (also Lindelöf) is equivalent with second countability, then every subspace of a separable metric space is separable (and Lindelöf, for that matter). –  Thomas E. Aug 14 '12 at 4:20
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