Re: Analysis(Topology) with connected.
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 23 Dec 2007 03:40:42 -0500
On Sun, 23 Dec 2007 00:18:55 -0800 (PST), mina_world@xxxxxxxxxxx
wrote:
On 12?23?, ??4?04?, quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 22:39:14 -0800 (PST), mina_wo...@xxxxxxxxxxx
wrote:
On 12¿ù23ÀÏ, ¿ÀÀü7½Ã58ºÐ, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article <avuqm3p1p7igu2ll8sj4a9ennh8jf9s...@xxxxxxx>,
quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 11:10:28 -0800, The World Wide Wade
<aderamey.a...@xxxxxxxxxxx> wrote:
In article <1fjqm31uum19uf0da9g77e6u3760kbg...@xxxxxxx>,
quasi <qu...@xxxxxxxx> wrote:
On Sun, 23 Dec 2007 01:44:11 +0900, "mina_world"
<mina_wo...@xxxxxxxxxxx> wrote:
Hello sir~
Def)
A set M subset R^n is "differentiably connected"
if every two point x, y in M subsetR^n can be joined
by a differentiable arc r : [0,1] ->R^n.
Give an example of a path-connected set that is not differentiably
connected.
The union of 2 distinct intersecting straight lines in R^2.
quasi
You can certainly connect any two points in your example with a 1-1
differentiable path. What is a differentiable arc?
I was assuming that a differentiable arc meant "smooth".
quasi
You can make it as smooth as you like: for example, f(t) =
(exp(-1/t^2), exp(-1/t^2)) for t > 0, (-exp(-1/t^2), exp(-1/t^2)) for
t < 0, f(0) = (0,0)
Why is this smooth ?
(0,0) is sharp point as y=|x|.
so, this is not smooth
and not differentable at (0,0)
so, this is an example of a path-connected set
that is not differentiably connected.
No ?
Def)
A path r : [a,b] -> R^n is continuously differentiable
if it is continuous,
r' is defined on (a,b)
and continuous on that open interval.
What's your definition of differentiable arc(path) ?
We were asking you.
With that definition, the set {(x,y) in R^2 | y = abs(x)} _is_
differentiably connected.
See The World Wide Wade's construction.
Oh, my god. Yes...
f(t) =
(exp(-1/t^2), exp(-1/t^2)) for t > 0,
(-exp(-1/t^2), exp(-1/t^2)) for t < 0,
f(0) = (0,0).
{(x,y) in R^2 | y = abs(x)} is the image of a 1-1 C^oo
function defined on R.
It's really true.
Def)
A path r : [a,b] -> R^n is continuously differentiable
if it is continuous,
r' is defined on (a,b)
and continuous on that open interval.
A path is piecewise continuously differentiable
if it is the concatenation of finitely many
continuously differentiable paths.
Theorem)
The following are equivalent
for an open subset D subset R^2.
1. D is connected.
2. D is path-connected.
3. D is differentiabley path-connected.
pf)
(1)->(2)
For a in D,
Let [a] be the collection of points b in D
such that there is some path connecting a to b.
By concatenation,
if b in [a], then [b]=[a],
so, {[a] | a in D} is a partition of D.
Fix a in D and take some c in [a].
There is some open ball U around c
such that U subset D.
and it is easy to get a path(even a line)
connecting c to every point of U.
Hence U subset [c] = [a].
This shows that [a] is open.
Now suppose that the theorem fails,
i.e. the partition has more than one element.
Pick a in D and
Let A = [a], B = Union {|b| | b not in [a]}.
By assumption, A is non-empty.
However,
A, B is a partition of D into disjoint open sets,
contrary to the assumption that D is connected.
(3)->(2) is immediate.
(2)->(1) is a basic theorem by topology.
(1)->(3) is proven exactly like (1)->(2),
letting [a] there be the collection of points
which can be connected to a
by a piecewise continuously differentiable path;
the proof that every [a] is open is the same.
-----------------------------------------
Give an example of a path-connected set that is not differentiably
connected.
Maybe, this is not false problem.
Let p be a 1-1, continuous function from [0,1] --> R^n such that the
arc length of the image of p is undefined.
Let M be the image of p.
It's immediate that M is path connected.
But since the arc length of M is undefined, M is not differentiably
connected.
quasi
.
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