Prove that in R^n, path components of open sets are open Solution We first prove

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We first prove that R^n is locally path connected. Definition. A space X is locally path connected if for every point x in X, for every neighborhood U of x, there is a path-connected neighborhood V of x such that V is contained in U. Let x be a point in R^n and let U be a neighborhood of x. In the standard topology of R^n, we can use -balls as a basis. Let V = B(x,) be a basis element surrounding x that is contained in U. V is path-connected: let a,b be points in V. Then the straight-line path f:[0,1]-->R^n defined by f(t) = (1 - t)a + tb is a path connecting a and b that lies entirely within B(x,). Now we have shown that for every x in R^n, for every neighborhood U there is a path-connected neighborhood V -- the appropriate -ball -- of x such that V is contained in U. Therefore R^n is locally path-connected. Now we show that if a space X is locally path-connected, it means that path components of open sets are open. Let X be a locally path-connected space; let U be an open set in X; let V be a path component of U. Since X is locally path-connected, if x is a point in V we can choose an open neighborhood of x, call it W, such that W is path-connected and W is contained in U. But since V is a path component of U and x is in V, W must be contained in V as well (recall that if two points are path-connected they must be in the same path component, since the path components are the equivalence classes of path-connected points -- so if W is path-connected and contains x, it must be contained in the path component of x, which is V). To sum up where we are, we now have x an element of W contained in V contained in U where W is a path-connected open neighborhood of x, V is a path component of U, and U is open in X. Recall that to prove a set is open, all we need to show is that for any element x in the set, there is an open set containing x that is contained in the set. Since we just showed that for any path component V in U, for any x in V there is an open set W such that x is in W and W is contained in V, that is sufficient to prove V is open. Since V was an arbitrary path component of U, and since U was an arbitrary open set in any locally path-connected X, we just proved that the path components of open sets are open (for any locally path-connected space). Since R^n is locally path-connected, as proven above, this completes the proof that path components of open sets in R^n are open. If you have the "local path-connectedness of spaces implies path components of open sets are open" concept as a theorem already, it suffices to prove that R^n is locally path-connected -- then you can jump straight to the conclusion.

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