Confused by Advanced Calculus Proof? Get Help Here!

[kingwinner]

http://www.geocities.com/asdfasdf23135/advcal18.JPG

Well, I don't get the idea of the proof at all...

I have no clue why they can let U be an open ball. Not all open sets are balls, and if they let U be an open ball, it doesn't not seem to me that the proof has covered ALL possibilities of open sets U with the given property.

Can someone please explain?

 

[AsianSensationK]

If I'm recalling correctly, it's because all open sets can be described as arbitrary unions of open balls. These are the basis elements. Open balls form a basis for the metric space, so you only need to check these and not every possible kind of open set.

Then again, I may be confusing topological and metric concepts here. The thing to remember is, you can describe pretty much any open set in R^k as a union of open balls.

 

[Dick]

http://www.geocities.com/asdfasdf23135/advcal18.JPG

Well, I don't get the idea of the proof at all...

I have no clue why they can let U be an open ball. Not all open sets are balls, and if they let U be an open ball, it doesn't not seem to me that the proof has covered ALL possibilities of open sets U with the given property.

Can someone please explain?

It says for ANY open set U. Balls ARE open sets. So in particular, the premise holds for them.

 

[kingwinner]

It says for ANY open set U. Balls ARE open sets. So in particular, the premise holds for them.

Yes, it says ANY, so we have to prove the statement for ANY open set that has the given property...

 

[CompuChip]

The assumption is the definition of continuous in topology. What you want to prove is what is usually taken as definition of continuity in analysis. The point of the proof is to show that if a function is "topologically continuous" then it is also "analytically continuous". So assuming f is "topologically continuous" - that is, the pre-image of any open set is again open - you have to prove that it is "analytically continuous", that is:
$$\forall \epsilon > 0 \exists \delta > 0: ||x - y|| < \delta \implies ||f(x) - f(y)|| < \epsilon$$

You can also prove the converse (then, assuming that it satisfies the epsilon-delta thing, which is basically the topological definition for open balls only, you do have to show that the pre-image of any open set is open), and then you have proven that the two definitions are equivalent for functions between real vector spaces (or, in other words, that when working on |R^n you can use the practical epsilon-delta definition instead of the rather abstract topological definition to prove continuity of a function).

 

[HallsofIvy]

Yes, it says ANY, so we have to prove the statement for ANY open set that has the given property...

NO! That is part of the hypothesis, not the conclusion. You use the hypothesis, you don't have to prove it. You don't have prove it "for any open set", you only have to prove "f is continuous on Rⁿ". You can use whatever open sets you want. In particular, if balls are sufficient, that is all you need.