Analysis Test Questions (did I get these?)

[futurebird]

Here are some test questions and my responses, did I get these right? These are the ones that I'm least certain about.

1. Show that in every metric space, finite sets are always closed.
   
   
   Finite sets contain no limit points. Hence every finite set contains all if its limit points.​
2. Suppose that \{ N_{a} \} is a collection of disjoint neighborhoods in \mathbb{R}^{2}. Show that the collection is at most countable.
   
   
   My proof was a little long but it went roughly like this:
   Because the neighborhoods are disjoint and because the rationals are dense in the reals it is possible to locate a point (a, b) where a and b are rational numbers in each neighborhood that is in no other neighborhood. So, we have a 1-1 correspondence between a subset of the countable set \mathbb{Q} \times \mathbb{Q} and the neighborhoods in \{ N_{a} \}. Hence, the set is at most countable. (Looking back, I can see I was thinking of pairwise disjoint, which we never defined... Rudin only mentions that two sets can be disjoint. http://en.wikipedia.org/wiki/Disjoint_sets" makes a distinction between pairwise disjoint and sets that merely have a disjoint intersection. I think a variation of this proof might still work, so can I even hope for partial credit?? Maybe?)​
3. Let p be a point in an arbitrary metric space X. Let A=\bigcap_{r \in \mathbb{R} >0} N_{r}(p) the intersection of all neighborhoods of p. Prove that A={p}.
   
   
   I said that a nested intersection of open sets could not be empty, and p is the only element contained in every set hence the intersection must be p.​
4. For a metric space X and a function f:X \rightarrow \mathbb{R} define Z(f) as the set of all x in X such that f(x) = 0. Show that if f is continuous, then Z(f) is closed.
   
   
   I said that {0} is a closed set in \mathbb{R} because it is finite. Hence f^{-1}(\{ 0 \}) must be a closed set in X because f is continuous.​

If you can let me know if any of these seem right or wrong it be a big help! I'm freaking out about my grade.

 

[futurebird]

No. 1 should read: "Finite sets contain no limit points. Hence every finite set contains all of its limit points."

 

[futurebird]

Maybe I should move this post to the analysis forum?

 

[Dick]

The only answer that's wrong big time is 3). An intersection of open sets CAN be empty, try the intersection of the intervals (0,1/n), and you can't use that p is the only point common to all the neighborhoods in the proof, that what you are trying to prove. For 2) Rudin clearly means pairwise disjoint. Otherwise it wouldn't be true.

 

[futurebird]

The only answer that's wrong big time is 3). An intersection of open sets CAN be empty, try the intersection of the intervals (0,1/n), and you can't use that p is the only point common to all the neighborhoods in the proof, that what you are trying to prove. For 2) Rudin clearly means pairwise disjoint. Otherwise it wouldn't be true.

An intersection of nested open neighborhoods can't be empty, though, right?

 

[Dick]

'nested' usually just means the (n+1)th neighborhood is contained in nth neighborhood. So the example I gave is nested. The set of the intersection of all neighborhoods of p isn't empty, since as you said, they all contain p. Your job is to show that the intersection is ONLY p.

 

[futurebird]

'nested' usually just means the (n+1)th neighborhood is contained in nth neighborhood. So the example I gave is nested. The set of the intersection of all neighborhoods of p isn't empty, since as you said, they all contain p. Your job is to show that the intersection is ONLY p.

Ah crud. Would it have been enough to show that given any point in a neighborhood of P there is a set smaller than that neighborhood that excludes it? Or is there more to it than that?

 

[Dick]

Ah crud. Would it have been enough to show that given any point in a neighborhood of P there is a set smaller than that neighborhood that excludes it? Or is there more to it than that?

That's it. If there is a point q in the intersection that's not equal to p, construct a neighborhood that excludes it. Proof by contradiction.