Completing Warsaw circle (S205)

[felixpernegger]

I'm back!

Admittedly, my proof for fixed point property is not 100% formal, but it is hard to formulate this properly, as we do not have a natural projection onto some coordinate like with the (some) other sin(1/x) curves.

I also suggest to look at the Knaster-Kuratowski fans (S125, S126) as (as far as I can see) this are the only embeddable into $\mathbb{R}^n$ spaces which are somewhat natural and not yet completed,

[prabau]

P89: let me see if I understand this argument, or at least modify it to something that seems slightly easier to me.

< order on $X$: ok. One can wrap the semi-open interval interval $J=[0,\infty)$ onto $X$ to get a continuous bijection $h:J\to X$ that also satisfies $s<t\implies h(s)<h(t)$ (in $X$).

Suppose $f:X\to X$ is continuous. By examining path-connected components of nbhds, one can see that the composition of functions $g=h^{-1}\circ f\circ h:J\to J$ is continuous.
Then argue as you did: if $f$ has no fixed point, neither does $g$. So $g(0)>0$. By continuity of $g$, $g(s)>s$ for all $s\in J$.

On the other hand, consider the point $x=(0,-1)\in $X$. Suppose $x=h(t)$. We have $g(t)>t$, which implies that there is a small nbhd $V$ of $f(x)$ in $X$ that consists of a single arc around $f(x)$, with bounded values of $h^{-1}$. (That's important for the rest of the argument.)

Now there is a sequence of points $x_n\in X$ with $x_n\to x$ and $h^{-1}(x_n)=t_n\to\infty$. By taking $n$ large enough, we'll have $f(x_n)\in V$, hence necessarily $g(t_n)<t_n$, which is a contradiction.

What do you think?

[prabau]

S205: the README mentions that there are some non-isomorphic variants of the space with the same name. I have added a link to some pictures to illustrate.

[prabau]

@felixpernegger not sure if you have seen the comment above #1535 (comment)

[felixpernegger]

P89: let me see if I understand this argument, or at least modify it to something that seems slightly easier to me.

< order on X : ok. One can wrap the semi-open interval interval J = [ 0 , ∞ ) onto X to get a continuous bijection h : J → X that also satisfies s < t ⟹ h ( s ) < h ( t ) (in X ).

Suppose f : X → X is continuous. By examining path-connected components of nbhds, one can see that the composition of functions g = h − 1 ∘ f ∘ h : J → J is continuous. Then argue as you did: if f has no fixed point, neither does g . So g ( 0 ) > 0 . By continuity of g , g ( s ) > s for all s ∈ J .

On the other hand, consider the point $x=(0,-1)\in X . Suppose x = h ( t ) . We have g ( t ) > t , which implies that there is a small nbhd V of f ( x ) in X that consists of a single arc around f ( x ) , with bounded values of h − 1 . (That's important for the rest of the argument.)

Now there is a sequence of points x n ∈ X with x n → x and h − 1 ( x n ) = t n → ∞ . By taking n large enough, we'll have f ( x n ) ∈ V , hence necessarily g ( t n ) < t n , which is a contradiction.

What do you think?

To be honest this might be a bit more formal but probably harder to follow along and still handwaves $g$ continuous. I don't think it matters all to much. I suppose the argument is the easiest to follow by taking a paper.

[yhx-12243]

I think we can currently approve this and (possibly) add more details in mathse.

[prabau]

would prefer to approve this after I can understand the proof in the mathse post

[prabau]

To be honest this might be a bit more formal but probably harder to follow along and still handwaves g continuous.

I think my proof had not too much handwaving about the continuity of $g$. But the rest of the proof is still not clear to me. Hopefully you can clarify in the mathse post.

[yhx-12243]

Picture storage for mse post.

Check https://math.stackexchange.com/questions/5116689. @prabau