On Sun, 4 May 2008, amu wrote:
> is this proof write to above question
>
I read only what is posted. Thus I don't know what you're proving.
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.
U is path connected subset
> 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.
Ok.
> 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.
> 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.
Your reasoning is wrong. Though it true that the closure of a connected
set is connected, it is not true that the closure of a path connected set
is path connected. For example, the topologists sin curve, the clusure
within R^2 of { (x,1/x) | 0 < x }.
> 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.
>
Multi-point zero dimensional spaces are not path connected
and some n dimensional spaces are not metric spaces, hence
don't have any balls.
The key to proving connected, locally path connected implies path
connected is the chain rule:
If C is an open cover of a connected space S, then for all a,b in S,
there is an (finite) over lapping chain from a to b of sets from C.
That is, some U1,.. U_n in C with a in U1, b in U_n,
for j = 1,.. n-1, Uj /\ U_(j+1) nonnul.
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