Thread: Proving connected/compact
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Old November 28th, 2008, 10:03 AM
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Originally Posted by GB89 View Post
Let D be a subset of R^n and f mapping from D to R^m be continuous.
a.) Prove that if D is connected, then the range of f is connected.
If U is an open subset of R^m then f^{-1}(U) is open in D. So if U, V are open and disjoint, and U\cup V covers the range of f, then f^{-1}(U), f^{-1}(V) are disjoint open sets whose union is D. You take it from there... .

Quote:
Originally Posted by GB89 View Post
b.) Prove that if D is compact, then the range of f is compact.
Similar idea. If you have a covering of the range of f by open sets, then their inverse images form an open cover of D, which has a finite subcover... .
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