Re: -- ties and quasi-paths
- From: William Elliot <marsh@xxxxxxxxxxxxxxxxxx>
- Date: Sat, 9 Feb 2008 04:31:09 -0800
On Sat, 9 Feb 2008, quasi wrote:
<marsh@xxxxxxxxxxxxxxxxxx> wrote:
f in C(L,S) is a long path from a to b when
L compact connected linear order
a = f(min L), b = f(max L)
f is long arc from a to b when
f is an injective long path from a to b
H in C(XxL,Y) is a long homotopy between f,g:X -> Y when
L compact connected linear order
for all x, H(x,min L) = f(x), H(x,max L) = g(x)
P is a quasi-path when P is the image of a long path.
P is a simple quasi-path when P is the image of a long arc.
Other expressions could be trail and road; q-path, simple q-path;
long path image, long arc image.
T tie between a and b, an a,b tie when a,b in T, T connected.
T ring at a when a in T, T connected, T not unicoherent
S is unicoherent when S connected, for all closed K,L,
K \/ L = S ==> K /\ L connected.
Notice that the circle S^1, is not unicoherent while R^2 and S^2 are.
Of interest is to study the implications between simply connected
and unicoherent. These parallels are to be considered
long path tie
long arc minimal tie
loop ring
A space S is long contractible when S is long homotopic to a point.
A space S is long simply connected when S connected and all rings
are long contractible.
There is a significant difference between
compact minimal tie
and
minimal compact tie.
I'll have to think about this.In the second part of that post, I introduced the notion of minimal tie
I want to review some properties about ordered topological spaces.
connected spaces, a category between connected and path connected.
What do you think of it?
My concern is whether
S is simply connected iff S unicoherent path connected.
S simply connected iff S path connected,
for all a, p loop at a, p homotopic p_a relative to { 0,1 }
H in C([0,1]^2,S) homotopy between
p,p_a:[0,1] -> S relative to { 0,1 } iff
for all r in [0,1], H(r,0) = p(r), H(r,1) = a
for all t in [0,1], H(0,t) = a = H(1,t)
Now for our mutual amusement, a variant of the topologist's sin curve
S = cl { (x, sin 1/x) | x in (0,1] }.
Add a simple curve from (1, sin 1) not intersecting S
that ends at (0,1). Oh boy, it's path connected.
Not only that, it's vacuously simply connected because
no loop can ever get around the whole dang contraption.
So there it is, a mutant freak,
simply connected, not unicoherent,
Well, that much has been unsolved. What's left to consider is
if path connected S is unicoherent, is S simply connected?
Anyway, this strengthens the notion that not unicoherent tells
better than simply connected, if there's a hole in my space.
In addition it's a ring, a minimal compact ring but not a minimal
ring. Removing {0}x[-1,1), leaves a not compact minimal ring.
The minimal compact ring has an inside and an outside.
The not compact minimal ring has inside and outside connected.
I'll wager it's not minimal tie connected.
Remember the two connected sets, based upon the topologist sin curve
that passed through each other? One was minimal tie connected, the other
was not.
Unlike S, S - {(0,0)} is not minimal tie connected. Thus remains the
nagging notion, do ties of a compact Hausdorff space have minimal ties?
----
.
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