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Question

Is $f(x)=(2x-1)/3$ a homeomorphism on the $2$-adic integers $\Bbb Z_2$? And does it expand or contract?

My attempt

$\lvert(2x-1)/3-(2y-1)/3\rvert_2=\lvert(2x-2y)\rvert_2=\frac12\lvert x-y\rvert_2$ so I'm fine on it preserving the topology, it just shrinks stuff.

In particular, I'm asking about the fact it doesn't surject. It maps from $\Bbb Z_2\to\Bbb Z_2^\times$

Also, I'm confused because it shrinks the pair $x,y$ but $\Bbb Z_2\to\Bbb Z_2^\times$ looks like it's growing stuff because it would take a ball of radius $1/2$ around zero to a ball of radius one.

I'm probably making a schoolboy error, please point it out. Or is there some paradoxical property regarding shrinking balls and their elements moving further apart?

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  • Do you mean the $2$-adic integers, as topological space? For context, see this question. – Dietrich Burde Nov 16 '22 at 10:57
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    I don't understand the problem. Doesn't this map the ball of radius one centered at $x=0$ to the ball of radius $1/2$ centered at $x=-1/3$? Just what you would expect from a shrinking map. Not that $2$-adic balls would have a unique center, but that's not the key issue here. – Jyrki Lahtonen Nov 16 '22 at 11:13
  • thanks @JyrkiLahtonen although you failed to work out my error, that does clear it up. I was thinking of the ball of radius one around zero, but of course zero is not in the image of $f$. Just the main homeomorphism question left. – it's a hire car baby Nov 16 '22 at 11:42
  • @DietrichBurde yes I do. I'm not sure how the linked question helps though. – it's a hire car baby Nov 16 '22 at 11:45
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    But have you not answered the question of homeomorphism yourself? It is clear that $\frac{2x-1}{3}$ is a $p$-adic unit for any $x\in\mathbb{Z}_2$, so it's clearly not surjective. – SomeCallMeTim Nov 16 '22 at 11:45
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    You observed that $0$ is not in the image, so it isn't surjective, hence not a homeomorphism. – Jyrki Lahtonen Nov 16 '22 at 11:46
  • @SomeCallMeTim if there were a theorem saying that homeomorphisms must surject then this would indeed answer the question. But I do not have that theorem at the moment. – it's a hire car baby Nov 16 '22 at 11:47
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    @samerivertwice What is your definition of a homeomorphism? The usual one is a continuous bijective function with a continuous inverse. – SomeCallMeTim Nov 16 '22 at 11:48
  • @JyrkiLahtonen sorry for the trivial error. Doing my best. – it's a hire car baby Nov 16 '22 at 11:53
  • @SomeCallMeTim thanks for your help. tbh, I was thinking of bijective as needing a well-defined inverse only for elements which are in the image which was kind of an intuitive idea, but it's clear now. – it's a hire car baby Nov 16 '22 at 11:56
  • "I'm not sure how the linked question helps though." - The linked question shows that you have worked on this and what your background is. This is valuable to know for people, so that they can give an answer, which is helpful to you. (the topic is very similar, even the same map was considered there, too, named $g(x)$). – Dietrich Burde Nov 16 '22 at 12:01
  • ah ok @DietrichBurde it helps people thinking of answering or commenting. Thanks. – it's a hire car baby Nov 16 '22 at 12:03

1 Answers1

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A homeomorphism needs to biject by definition so $f$ trivially isn't a homeomorphism.

And it contracts in the sense that the smallest ball in which one can contain the range is smaller than the smallest ball containing the domain.

Because however, $f$ is a homeomorphism onto its image, it is called a topological embedding.

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