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.
> 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.
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