Proving connected/compact

GB89 · #1 $Old$ November 28th, 2008, 10:20 AM

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.)

Opalg · #2 $Old$ November 28th, 2008, 11:03 AM

Quote:

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... .

GB89 · #3 $Old$ November 29th, 2008, 08:53 PM

Quote:

Originally Posted by Opalg $View Post$

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... .

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... .

For the first part, would this be correct from where you stopped?

..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.

Opalg · #4 $Old$ November 30th, 2008, 02:11 PM

Quote:

Originally Posted by GB89 $View Post$

For the first part, would this be correct from where you stopped?

..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.

That is the right idea, except that I never asked that U and V should be nonempty. You can either add that requirement at the beginning of the proof, or you can say that since D is connected it follows $f^{-1}(U)$ or $f^{-1}(V)$ must be empty and hence either U or V must be nonempty. Either way, you arrive at the conclusion that the range of f cannot be the union of two nonempty disjoint open subsets.

Quote:

Originally Posted by GB89 $View Post$

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.

I'm using the definition of compactness that says that D is compact if every open covering of D has a finite subcover. More explicitly, if $\{U_i\}$ is a collection of open subsets of D whose union is the whole of D then there exist finitely many of these sets, say $U_1,U_2,\ldots,U_n$ such that $\bigcup_{j=1}^nU_j=D$ .

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 $\{V_i\}$ is a collection of open subsets of the range of f whose union is the whole of the range. Then $\{f^{-1}(V_i)\}$ is a collection of open subsets of D whose union is the whole of D. Therefore there is a finite subset $V_1,V_2,\ldots,V_n$ whose union is the whole of D, and it follows that $\bigcup_{j=1}^nV_j$ 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.