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September 28th, 2015, 09:05 AM  #1 
Newbie Joined: Sep 2015 From: São Paulo, Brazil Posts: 2 Thanks: 1 
The following is an excerpt from Serge Lang's "Real and Functional Analysis". In the proof, the author defines $\left \{k...k_n \right \}$ as a subset of D. How does he know that D is big enough to contain a set of elements that can be indexed to n? I assume that, by n, the textbook means an index set with the same cardinality as the natural numbers. Last edited by skipjack; September 28th, 2015 at 10:55 AM. 
September 28th, 2015, 01:18 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,134 Thanks: 621 
n is a finite number used in the induction.

October 4th, 2015, 01:16 PM  #3 
Newbie Joined: Sep 2015 From: São Paulo, Brazil Posts: 2 Thanks: 1 
@maschke How come $\displaystyle n$ is finite if its being used in a denumeration? To denumerate a sequence means "to trace a function from $\displaystyle Z^+$ to the set of the sequence terms". $\displaystyle n$ should be at least as big as $\displaystyle Z^+$... 
October 4th, 2015, 08:55 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,134 Thanks: 621 
It's an induction. Step 1. D is infinite, therefore it's nonempty, therefore by the wellordering property of the positive integers, it has a smallest element. Call that k1. Note that Lang has left out that level of detail but it's important to realize that you need wellordering in order to know that the smallest element of D exists. Step 2. D \ {k1} is nonempty (because D is infinite and {k1} is finite) so it has a smallest element. Call it k2. Step 3. D \ {k1, k2} is nonempty (because D is infinite and {k1, k2} is finite) so it has a smallest element. Call it k3. Step 4. D \ {k1, k2, k3} is nonempty (because D is infinite and {k1, k2, k3} is finite) so it has a smallest element. Call it k4. etc. Does that make the structure of the proof more clear? It's an induction on n. You keep picking elements from D and adding them to the ksequence. At each step of the induction, {k1, k2, ..., kn} is a finite set, and D \ {k1, k2, ..., kn} is nonempty, so you can always pick k_(n+1). Last edited by Maschke; October 4th, 2015 at 09:02 PM. 
October 4th, 2015, 09:09 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119 
If D is not denumerable, neither is Z+. Contradiction.

October 5th, 2015, 09:47 AM  #6 
Senior Member Joined: Aug 2012 Posts: 2,134 Thanks: 621  
October 5th, 2015, 10:29 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119 
Begin enumerating Z+ by enumerating D.

October 5th, 2015, 02:01 PM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119 
An infinite set of anything is countable by definition. How else would you know it's infinite?

October 5th, 2015, 08:38 PM  #9  
Senior Member Joined: Aug 2012 Posts: 2,134 Thanks: 621  Quote:
The theorem is proving the statement that every infinite subset of Z+ is countable. You can't assume the thing you're trying to prove. By the way I'm not familiar with this text but I know Lang's graduate algebra book. He can be terse. You need to mentally add in all the details he's skipping over. It pays to read Lang slowly. Last edited by Maschke; October 5th, 2015 at 08:45 PM.  
October 6th, 2015, 12:17 PM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119  Induction
Maschke, Thanks. You are clear and correct on all counts. But just for fun: Induction Principle: If Pn > (IMPLIES) Pn+1 and P1=0, all Pn true. A subset of D does not have a lowest member because I removed its lowest member. It's a postulate (well ordering). So the OP is true by postulate. But I really don't believe that. If you accept the postulate, anything you prove after that is a valid theorem. And if b is true whenever a is true, a > b even though there is no causality. Like I said, just for fun. 

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denumerability, denumerable, subset 
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