Proving Subdivision and Refinement: Is it as Simple as it Seems?

[TheyCallMeMini]

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.

Homework Statement

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.

Homework 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.

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.

 

[mfb]

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]

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]

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]

I didn't see the drop down menu for the element, union, and implication arrows. Where are those?

 

[mfb]

They are written with the -tags and LaTeX codes. See the <a href="https://www.physicsforums.com/showpost.php?p=3977517&amp;postcount=3" class="link link--internal">FAQ entry</a> for details, or quote my post to see its code.