Example of a Homeomorphism f: Q -> Q not order preseving or order-reversing

[GridironCPJ]

Any ideas? I really can't think of any myself, as I'm quite the amatuer at topology.

 

[Bacle2]

I know this may be little more than redefining the problem, but, how about thinking of a homeo of the reals to themselves with your wanted property, and then restricting to Q. This does not guarantee, but may help.

Problem I see is that for a map from R to itself to be invertible, it has to be monotone.

 

[GridironCPJ]

I've considered thinking of it this way, but I didn't come up with anything. I get the idea that this must be excusive to the rationals or irrationals, since they are both dense, but every interval contains points not in the set.

 

[dodo]

I'm not sure if I understood the problem correctly but, wouldn't a continuous and monotone descending function do?

 

[GridironCPJ]

No, that would simply reverse the order. Like I said, I think the function needed is something that acts "special" with the rationals.

 

[dodo]

Ah, you mean "neither order-preserving nor order-reversing". I read (not order-preserving), or order-reversing.

Perhaps you want to change the topology from the Euclidean one. For example, any function from Q to Q with the discrete topology (on the domain is enough) will be continuous; and, as it can be any function, you could choose any weird bijection to your liking. My 2 cents (the only I have left in my pocket :).

 

[lavinia]

- for the extended rationals - the rationals with infinity added - inversion is a non-monotone homeomorphism.

- If you do not allow the point at infinity, I would try to show that if a Cauchy sequence is mapped to a Cauchy sequence then all equivalent Cauchy sequences are mapped to equivalent Cauchy sequences. This would mean that the map is extendable to the reals.

 

[Bacle2]

lavinia:

If you're thinking 1-pt-compactification, then your resulting space would not be Hausdorff, since Q is not locally-compact (e.g., the sequence 1, 1.4, 1.414,... has no convergent subsequence).

 

[lavinia]

lavinia:

If you're thinking 1-pt-compactification, then your resulting space would not be Hausdorff, since Q is not locally-compact (e.g., the sequence 1, 1.4, 1.414,... has no convergent subsequence).

OK. I was just thinking of the image of the rationals in the circle under inverse stereographic projection then adding the point at infinity and taking the subset topology. That doesn't work?

 

[Citan Uzuki]

It's possible to show a homeomorphism exists from $\mathbb{Q}$ to $\mathbb{Q}$ that is not monotone, but difficult to describe it explicitly (although one could create an explicit formula in principle if pressed). The key is in the following fact: any two nonempty countable densely ordered sets without endpoints are order-isomorphic. So you could let $A = \{x\in \mathbb{Q} : x<\sqrt{2}\}$, $B = \{x \in \mathbb{Q} : x>\sqrt{2}\}$, and then let $f:A \rightarrow B$ be an order-preserving bijection from A to B. Note that since the metric topology on the rationals agrees with the order topology, f is in fact a homeomorphism from A to B. So then let $G:\mathbb{Q} \rightarrow \mathbb{Q}$ be given by $g(x) = f(x)$ if $x\in A$ and $g(x) = f^{-1}(x)$ if $x\in B$. Then the restriction of g to either A or B is continuous, and since A and B are both open in $\mathbb{Q}$, g is continuous. And we also have that $g^{-1} = g$, so g is in fact a homeomorphism, which is neither order-preserving nor order-reversing.

 

[lavinia]

It's possible to show a homeomorphism exists from $\mathbb{Q}$ to $\mathbb{Q}$ that is not monotone, but difficult to describe it explicitly (although one could create an explicit formula in principle if pressed). The key is in the following fact: any two nonempty countable densely ordered sets without endpoints are order-isomorphic. So you could let $A = \{x\in \mathbb{Q} : x<\sqrt{2}\}$, $B = \{x \in \mathbb{Q} : x>\sqrt{2}\}$, and then let $f:A \rightarrow B$ be an order-preserving bijection from A to B. Note that since the metric topology on the rationals agrees with the order topology, f is in fact a homeomorphism from A to B. So then let $G:\mathbb{Q} \rightarrow \mathbb{Q}$ be given by $g(x) = f(x)$ if $x\in A$ and $g(x) = f^{-1}(x)$ if $x\in B$. Then the restriction of g to either A or B is continuous, and since A and B are both open in $\mathbb{Q}$, g is continuous. And we also have that $g^{-1} = g$, so g is in fact a homeomorphism, which is neither order-preserving nor order-reversing.

cool. so dividing the rationals into an arbitrary even number of intervals and pairwise shuffling them will work.

 

[Bacle2]

OK. I was just thinking of the image of the rationals in the circle under inverse stereographic projection then adding the point at infinity and taking the subset topology. That doesn't work?

But I think the topology of the compactification is kind-of messy, since (at least that I know) , there is no nice characterization of the compact subsets of Q (closed+bounded doesn't do it; convergent sequences are, but I can't see a nice way of characterizing them all to determine the topology in the compactification).