Question about proof of associative law for sets

[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

 

[yyat]

Hi pamparana,

What it comes down to is that "or" (http://en.wikipedia.org/wiki/Logical_disjunction" for
((p or q) or r)
and
(p or (q or r))
are the same.

 

[statdad]

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.

$$\begin{align*} x \in (A \cup B) \cup C & \leftrightarrow x \in (A \cup B) \text{ or } x \in C \\ & \leftrightarrow x \in A \text{ or } x \in B \text{ or } x \in C \\ & \leftrightarrow x \in A \text{ or } x \in (B \cup C) \\ & \leftrightarrow x \in A \cup (B \cup C) \end{align*}$$

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.