On Sun, 4 May 2008 22:43:52 -0700, William Elliot
<marsh@[EMAIL PROTECTED]
> wrote:
>On Sun, 4 May 2008, amu wrote:
>
>> is this right proof
>>
>A proof of what? Please don't remove essential context, it's im****tant.
>
>Please included essential content _within_ your reply
> http://oakroadsystems.com/genl/unice.htm
>
>> Let C be a connected set that is also locally path connected. Pick any
>> point x in C, and let U be the set of points in C that are path
>> connected to x. Thus U is a subset of C. Let y be a point in U.
Enclose
>> y in an open set H in C, such that y is path connected to all of H.
>> Since an arc can run from x to y to anything in H, H is in U. Therefore
>> U is the union of open sets and is open, relative to C.
>>
>Theorem.
> Connected and locally path connected S ==> S path connected.
>Use the chain rule.
What do you mean by "chain rule"?
(The phrase "chain rule" usually refers to the theorem/technique
for differentiating a function defined as the composition of
two functions...)
>> Let y be a point in C that is a limit point of U. Put an open set H
>> around y such that H is path connected. Let z be common to H and U. Now
>> x connects to z connects to y, and y is in U.
>>
>Not following because of loss of context.
>
>> Since U contains its limit points it is closed. thus U is both open and
>> closed in C. If U is not all of C, separate U and the rest of C in open
>> sets. This contradicts the fact that C is connected. Therefore U is all
>> of C, and C is path connected.
>>
>Limit points, are a calculus hang over.
>
>> In n dimensional space, every open ball is path connected, and every
>> open set is locally path connected, hence every open connected set is
>> path connected.
>>
>Not so in a zero-dimensional space.
>Give an example of a non-metrizable 1-dimensional space.
David C. Ullrich
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