# Question about proof of associative law for sets

1. Mar 21, 2009

### pamparana

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

2. Mar 21, 2009

### yyat

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.

3. Mar 21, 2009

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.

\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*}

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