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.
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.
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