On 5 Mai, 05:49, amu <amu78...@[EMAIL PROTECTED]
> wrote:
> is this right proof
>
> 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.
>
> 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.
I'd do this like: Let y in C\U and H a path-conncted open
neighbourhood of y. Then H and U are disjoint, for otherwise a path
from x to y would exist via a point in the intersection.
Hence U and C\U are both open, hence one of them must be empty.
Since x in U, it follows C\U={} and C=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.
>
> 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.
The proof is quite alright, but your final "A space that is connected
and is locally path-conncted is path-conncted". This does not show "If
X,Y are locally path-conncted thane a subset A of XxY is connected if
and only if it is path-connected". In fact, that claim is false.
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