I need help with this proof

[Cratylus]

Prove that a unit circle centred at the origin is homeomorphic to a square with a unit perimeter centred at the origin

l have researched it and looking at answers,it gives me headaches
So let C:={[MATH]x^{2}+y^{2}=[/MATH] 1 }. S={(0,1) x (0,1) }
Show C~S

l was thinking first to show each vertex of the square is homeomorphic to (0,0) on the circle
Then I found this: https://math.stackexchange.com/q/2765821/919044

(2) and (3) l can handle,Not sure about (1)

Any help would be greatly appreciated Dealing with the geometry gives me the heebie geebies.

 

[Dr.Peterson]

Why do you spend your time searching rather than trying to solve the problem yourself? What you found is not at all the same problem.

And your S is not a square, much less a square with unit perimeter centered at the origin.

Look for a mapping that takes each point on the square to a point on the circle.

 

[JeffM]

@Dr.Peterson

This question is not my bag at all, but doesn't the mapping have to be bijective?

 

[Cratylus]

@Dr.Peterson

This question is not my bag at all, but doesn't the mapping have to be bijective?

Yes it does

 

[Cratylus]

Why do you spend your time searching rather than trying to solve the problem yourself? What you found is not at all the same problem.

And your S is not a square, much less a square with unit perimeter centered at the origin.

Look for a mapping that takes each point on the square to a point on the circle.

To get an idea on how to solve it. I suck at geometry. I found bijection for circle to square
that I could use. But doesn’t the link give
me a good method to do it? If not explain why?
S={(0,0),(10),(0,1),(1,1)}

 

[lev888]

To get an idea on how to solve it. I suck at geometry. I found bijection for circle to square
that I could use. But doesn’t the link give
me a good method to do it? If not explain why?
S={(0,0),(10),(0,1),(1,1)}

The problem you linked deals with 2d shapes: disk, square. It's not clear whether your problem refers to 2d shapes or curves. If 2d, it matters whether the sets are open or closed.

 

[Dr.Peterson]

@Dr.Peterson

This question is not my bag at all, but doesn't the mapping have to be bijective?

Sure. and it will be. But the first step is to find one, and then either invert it (the inverse may be most easily stated in piecewise form) or just prove that it is one-to-one, without actually stating the inverse. And my suggestion was that it is easier to see by starting with the square.

To get an idea on how to solve it. I suck at geometry.

I am disturbed by the frequency with which students today resort to searching for ready-made answers rather than work harder on figuring out how to use what they've learned. Yes, sometimes you can get good ideas by searching; but that doesn't prepare you for taking tests or for doing work on new problems later. And quite often what you find is not what you think it is, or uses techniques unrelated to what you are learning, or otherwise is only a distraction.

I found bijection for circle to square
that I could use. But doesn’t the link give
me a good method to do it? If not explain why?

There may be something there that you could use, with modification to apply to the circle rather than the open disk. Give it a try.

But what is the bijection you found? That might be worth more discussion.

S={(0,0),(10),(0,1),(1,1)}

Does this say that the square consists of four points (ordered pairs)? That isn't a square, and certainly isn't the particular square they are talking about in the problem, whose center is (0,0).

 

[Cratylus]

I am self. studying it. I am 58 . I graduated years ago. As to the square in the problem, it just says u square centred at (0,0). So l assumed square of
ordered points . I don’t have much to go on
Sets ,metric properties continuity and def. of
homeomorphism Conover states, you will be
introduced to advanced ideas where you should
convince yourself they are true,without having to
prove them (If you can explain it to a layman, you
succeeded)Most of exercises in this section are
just that
The book I am using is a first course in Topology by Robert conover

well when l was in school I aced intro analysis. We used spivak’s Calcuks

 

[Dr.Peterson]

Conover states, you will be
introduced to advanced ideas where you should
convince yourself they are true,without having to
prove them (If you can explain it to a layman, you
succeeded)

In that case, maybe just drawing a picture will do (and if it isn't enough, it's an important start):



I've drawn a square with perimeter 8 rather than 1, but the idea would be similar. I think it's clear that we have a bijection between points on the square and points on the circle.

I also showed the equation of the square, though you don't need to use it; that's just a fun fact. In a proof, you'd probably want to treat the square as the union of four segments.

 

[JeffM]

Sure. and it will be. But the first step is to find one, and then either invert it (the inverse may be most easily stated in piecewise form) or just prove that it is one-to-one, without actually stating the inverse. And my suggestion was that it is easier to see by starting with the square.

Thank you.

 

[Cratylus]

In that case, maybe just drawing a picture will do (and if it isn't enough, it's an important start):

View attachment 27491

I've drawn a square with perimeter 8 rather than 1, but the idea would be similar. I think it's clear that we have a bijection between points on the square and points on the circle.

I also showed the equation of the square, though you don't need to use it; that's just a fun fact. In a proof, you'd probably want to treat the square as the union of four segments.

Thanks for the image. The following bijection was found on Quora

 

[Dr.Peterson]

Thanks for the image. The following bijection was found on Quora
View attachment 27497

Though it's a little hidden there, you'll observe that a complete description of the bijection would be piecewise (four cases, only one of which is shown here) as I suggested. The map from the square to the circle can be stated more easily, essentially giving the unit vector in the same direction. Polar coordinates work nicely for the circle, but less so for the square, and it's a little harder this way to see what the mapping does.

In the StackExchange link you provided, there was a Related link to https://math.stackexchange.com/questions/103660/homeomorphism-from-square-to-unit-circle?rq=1, which gives the mapping I had in mind, with an incorrect attempt at an inverse using absolute values.