# Are the rationals minus a point homeomorphic to the rationals?

A while ago I was dreaming up point-set topology exam questions, and this one came to mind:

Is $\mathbb Q\setminus \{0\}$ homeomorphic to $\mathbb Q$? (Where both sets have the subspace topology induced from the standard topology on $\mathbb R$.)

However, I couldn't figure this out at the time, and I'm curious to see whether anyone has a nice argument. I'm not even willing to take a guess as to whether they are or aren't homeomorphic.

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## 3 Answers

A well-known theorem of Cantor says that any two countable dense linear orderings without endpoints are isomorphic as linear orders. So in particular there is an order-preserving bijection between $\mathbb{Q}$ and $\mathbb{Q} - \{0\}$. This bijection will be a homeomorphism if you give each space the order topology, which is the standard topology inherited from $\mathbb{R}$.

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Thanks. Do you know a reference for Cantor's theorem? – Grumpy Parsnip Dec 10 '10 at 12:28

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– Carl Mummert Dec 10 '10 at 12:37
That's not as hard as I thought it would be. Cool. – Grumpy Parsnip Dec 10 '10 at 12:49

Since the technique proving the homeomorphism is useful in many other situations, it may be worth adding some details to Carl's answer:

Suppose $A,B$ are two countable, dense linear orders without end points. We show that they are isomorphic by building an isomorphism $f:A\to B$. This is done by what we call a back-and-forth argument. Say that $A=\{a_n\mid n\in{\mathbb N}\}$ and $B=\{b_n\mid n\in{\mathbb N}\}$. We build $f$ by stages. At the end of stage $2n$ we have ensured that $a_n\in{\rm dom}(f)$, and at the end of stage $2n+1$, we have ensured that $b_n\in{\rm ran}(f)$.

The construction is simple. Begin by picking any $b\in B$ and letting $f(a_0)=b$. This completes stage 0.

Then we do stage 1: If $b=b_0$ we are done and go to stage 2. If $b<b_0$, we pick an $a\in A$ and set $f(a)=b_0$. Of course, since $f$ is to be an isomorphism, we better ensure that $a_0<a$. But this is trivial to accomplish, since $A$ has no endpoints. Similarly, if $b_0<b$, then we pick $a$ so that $a<a_0$.

In general, at stage $2n$ do the following: If $a_n$ is already in the domain we have built, we are done with this stage. Otherwise, if $a_n$ is larger than all the elements in the domain of $f$ so far, pick an element $c$ of $B$ larger than all the elements in the range of $f$ so far and set $f(a_n)=c$; this is possible since $B$ has no largest element. If $a_n$ is smaller than all elements in the current domain of $f$, pick $c$ in $B$ smaller than all elements in the current range of $f$, and set $f(a_n)=c$. Again, this is possible, since $B$ has no smallest element. Finally, if $a_n$ is between elements of the current domain of $f$, pick $d,e$ in the current domain of $f$ so $d<a_n<e$, $d$ is largest below $a_n$, and $e$ is smallest above $a_n$. Then pick in $B$ some $c$ between $f(d)$ and $f(e)$ and set $f(a_n)=c$. This is possible, since $B$ is dense in itself. This completes this stage.

At stage $2n+1$ we do the same, but now ensuring that $b_n$ is put in the range of $f$.

This construction gives us an isomorphism $f$ at the end: Even stages ensure the domain of $f$ is all of $A$, odd stages that the range is all of $B$. The construction is designed so for $\alpha,\beta$ in the domain of $f$, $\alpha<\beta$ iff $f(\alpha)<f(\beta)$. But this is precisely what it means to be an isomorphism.

The method of back-and-forth is very flexible. For example, it shows that any two countable random graphs are isomorphic. There are plenty of applications of this technique.

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I'm a big fan of this argument in general, but for some reason, I especially like packaging it as the Rasiowa-Sikorski Lemma applied to the set of partially defined order preserving maps from $A$ to $B$. – Jason DeVito Dec 11 '10 at 0:17
Thanks for giving the argument here for easy reference. – Grumpy Parsnip Dec 11 '10 at 12:44
Hi Andres, I hope you don't mind that I added a missing \$. This answer was linked to me to answer one of my own questions. – yunone May 5 '11 at 21:01

The ordered approach is fine. A classical theorem by Sierpinski says that all countable metric spaces without isolated points are homeomorphic. Q is such a space. It also implies $\mathbf{Q} \setminus \{0\}$ is homeomorphic to $\mathbf{Q}$ and $\mathbf{Q} \times \mathbf{Q}$ e.g., or any finite product for that matter. A proof is at the topology atlas, topology explained

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Wow, that really defies intuition. – Grumpy Parsnip Dec 13 '10 at 8:50
Such characterizations are very nice, though. And in a way, it is intuitive, if you think longer about it. Other spaces that have such characterisations are R, the irrationals, the Cantor set, the Cantor set minus a point, to name the most famous ones.. – Henno Brandsma Dec 13 '10 at 21:51
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