Subdivisions/Refinement Proof


by TheyCallMeMini
Tags: proof
TheyCallMeMini
TheyCallMeMini is offline
#1
Feb11-13, 05:18 PM
P: 5
I understand this isn't a homework area but there is always so much more traffic in this forum rather than the homework. All I'm looking for is a clarification that my ideas to prove both parts are in fact accurate.


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

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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mfb
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#2
Feb11-13, 05:26 PM
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P: 10,791
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.
This is not a logical "or".

Consider a simple setup:
A={1}
B={2}
The statement "the set has exactly one element" is true for both A and B, but not for their union C={1,2}.

To show that E := D1 u D2 is a subdivision, you have to show that E is a finite subset of [a,b] and a and b are in E.

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...
If you prove "1." first, that works.
TheyCallMeMini
TheyCallMeMini is offline
#3
Feb11-13, 05:33 PM
P: 5
See I was thinking more along the lines if we had x an element of A then its simple enough to just say X is in A u B?

Because it doesn't matter if X is in B since we can just add another set, regardless its still in A. I'm just lost how to incorporate the union into the proof, I guess I'll spend more time on that.

Haha. See within 5 minutes I get a reply in this one and not the homework =P. Thank you!

mfb
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#4
Feb11-13, 05:37 PM
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P: 10,791

Subdivisions/Refinement Proof


Quote Quote by TheyCallMeMini View Post
See I was thinking more along the lines if we had x an element of A then its simple enough to just say X is in A u B?
##x \in A \Rightarrow x \in (A \cup B)##, and I think you can use this as it is very elementary.
TheyCallMeMini
TheyCallMeMini is offline
#5
Feb11-13, 05:48 PM
P: 5
I didn't see the drop down menu for the element, union, and implication arrows. Where are those?
mfb
mfb is offline
#6
Feb12-13, 09:23 AM
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P: 10,791
They are written with the [itex]-tags and LaTeX codes. See the FAQ entry for details, or quote my post to see its code.


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