I've been having trouble in my Advanced Calculus class and I'm not sure how to prove this. I was wondering if someone here could help.
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.
b.) Prove that if D is compact, then the range of f is compact.
Thanks in advance (and sorry I don't know how to get the math symbols on this site to make it clearer.)
..are disjoint open sets whose union is D. This would imply that D is not connected. But this is a contradiction. Thus U and V cannot be open and disjoint with their union covering the range of f, so the range of f is connected.
For part (b.), I'm really not sure what I'm supposed to do, even with your tip. I'd really appreciate it if you kept going on that one. Thanks so much.
Then the idea of the proof is to show that if D has that property then so does the range of f. So suppose that is a collection of open subsets of the range of f whose union is the whole of the range. Then is a collection of open subsets of D whose union is the whole of D. Therefore there is a finite subset whose union is the whole of D, and it follows that is the whole of the range of f.
Of course, if you are using another definition of compactness then you'll have to find another proof.