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Subdivisions/Refinement Proof

by TheyCallMeMini
Tags: proof
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TheyCallMeMini
#1
Feb11-13, 04:44 PM
P: 5
1. The problem statement, all variables and given/known data

If each of D1 and D2 is a subdivision of [a,b], then...
1. D1 u D2 is a subdivision of [a,b], and
2. D1 u D2 is a refinement of D1.


2. Relevant equations

**Definition 1: The statement that D is a subdivision of the interval [a,b] means...
1. D is a finite subset of [a,b], and
2. each of a and b belongs to D.


**Definition 2: The statement that K is a refinement of the subdivision D means...
1. K is a subdivision of [a,b], and
2. D is a subset of K.



3. The attempt at a solution

I just proved that, "If K is a refinement of H and H is a refinement of the subdivision D of [a,b], then K is a refinement of D." Well I havent wrote it down but the 2nd definition part 2 is what makes it easy to relate just being a transitive proof.

My problem is that I've taken a lot of logic courses in the past so when I see the union of two variables I only need to prove that one is actually true. In this particular situation both are true so its obvious but I don't know how to state that fact.

For the 2nd part of the proof, wouldn't I just say that D1 is a subset of itself, and its already given that D1 is a subdivision of [a,b]? It just seems too easy...


I also had questions about proofs I've already turned in that I did poorly on but I didn't want to flood this place with questions.
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ArcanaNoir
#2
Feb12-13, 09:08 AM
ArcanaNoir's Avatar
P: 753
Quote Quote by TheyCallMeMini View Post
For the 2nd part of the proof, wouldn't I just say that D1 is a subset of itself, and its already given that D1 is a subdivision of [a,b]? It just seems too easy...
Yes, I agree. Since by the first part you will have already shown that D1 u D2 is a subdivision of [a,b], all you have to do in part 2 is simply state that D1 is a subset of D1 u D2, thus the definition of refinement is satisfied.


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