Register to reply

Question about proof of associative law for sets

by pamparana
Tags: associative, proof, sets
Share this thread:
pamparana
#1
Mar21-09, 03:47 PM
P: 127
Hello,

Trying to go through Tom Apostle text on Calculus. There is an exercise about proving the associative law for sets:

So, (A U B) U C = A U (B U C)

So, if we assume x to be an element in set in left hand side, than we can say x belongs at least to either A, B or C which in turn means that x is also an element in set in right hand side and then we can say that the LHS and RHS are subsets of each other...

Is this a valid proof? I am never sure with these. It is really tricky to prove such ideas that we take for granted in every day life!

Anyway, I would be really grateful for any help you can give this old man.

/Luca
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
yyat
#2
Mar21-09, 04:04 PM
P: 321
Hi pamparana,

What it comes down to is that "or" (logical disjunction) is associative, so ((x in A or x in B) or (x in C)) is the same as (x in A or (x in B or x in C)). Either you take this for granted, or you check that that the truth tables for
((p or q) or r)
and
(p or (q or r))
are the same.
statdad
#3
Mar21-09, 08:31 PM
HW Helper
P: 1,361
Quote Quote by pamparana View Post
Hello,

Trying to go through Tom Apostle text on Calculus. There is an exercise about proving the associative law for sets:

So, (A U B) U C = A U (B U C)

So, if we assume x to be an element in set in left hand side, than we can say x belongs at least to either A, B or C which in turn means that x is also an element in set in right hand side and then we can say that the LHS and RHS are subsets of each other...

Is this a valid proof? I am never sure with these. It is really tricky to prove such ideas that we take for granted in every day life!

Anyway, I would be really grateful for any help you can give this old man.

/Luca
You are essentially correct. (The other post is correct too, but is really a round-a-bout way to assume exactly what you want to prove). You might see the proof of your statement organized formally this way.

[tex]
\begin{align*}
x \in (A \cup B) \cup C & \leftrightarrow x \in (A \cup B) \text{ or } x \in C \\
& \leftrightarrow x \in A \text{ or } x \in B \text{ or } x \in C \\
& \leftrightarrow x \in A \text{ or } x \in (B \cup C) \\
& \leftrightarrow x \in A \cup (B \cup C)
\end{align*}
[/tex]

I've use [tex] \leftrightarrow [/tex] to represent the phrase "if and only if" (I couldn't get the usual double arrow to work, sorry).
Hope this helps.


Register to reply

Related Discussions
Proof involving sets. NEED HELP Calculus & Beyond Homework 5
Proof about operations on sets Calculus & Beyond Homework 2
Proof of CLopen sets Calculus & Beyond Homework 2
More proof on sets help Set Theory, Logic, Probability, Statistics 2