Hello, I am having trouble proving the following theorem:

If A(1)....A(n) are finite pairwise disjoint sets then A(1) U A(2)U.....U A(n)= A(1)+....+A(n)

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- Mar 24th 2011, 01:03 PMqwerty10Addition Rule- Extended
Hello, I am having trouble proving the following theorem:

If A(1)....A(n) are finite pairwise disjoint sets then A(1) U A(2)U.....U A(n)= A(1)+....+A(n) - Mar 24th 2011, 01:21 PMPlato
I think that you have a good deal of explaining to do.

That is totally non-standard notation.

What do those symbols mean?

Does A(1) mean $\displaystyle A_1~?$

If so how does one read $\displaystyle A_1\cup A_2$=$\displaystyle A_1+ A_2~?$.

That does not make any sense. - Mar 24th 2011, 01:34 PMqwerty10
Yes that was my intended way of writing A(1), unable to do it. The theorem reads exactly what I have wrote above....if the sets are finite pairwise disjoint then the union of all the sets is equal to the addition of all the sets

- Mar 24th 2011, 01:44 PMPlato
Are you translating into English?

That makes no sense in mathematics.

There is a theorem about**counting**. The symbol $\displaystyle \|A\|$ stands for the number of elements is the finite set $\displaystyle A$.

If $\displaystyle A_n:n=1,2,\cdots,K$ is a collection of pairwise disjoint finite sets then $\displaystyle \displaystyle \left\| {\bigcup\limits_n {A_n } } \right\| = \sum\limits_{n = 1}^K {\left\| {A_n } \right\|} $.

Is that what is meant by the notation? - Mar 24th 2011, 01:49 PMqwerty10
The theorem is to do with counting so that must be the notation

- Mar 24th 2011, 01:56 PMPlato
- Mar 24th 2011, 02:01 PMqwerty10
That is where my problem lies in the induction proof:

I know to begin with the case n=2 so proving the basic counting principle.

Then I assume true for for n and take an n+1 element set and prove its true.

I am just struggling to write out a formal proof - Mar 24th 2011, 02:12 PMPlato
Suppose that $\displaystyle \{A_n:n=1,2,\cdots,K\}$ is a collection of pairwise disjoint finite sets and it is thue that $\displaystyle \displaystyle \left\| {\bigcup\limits_n {A_n } } \right\| = \sum\limits_{n = 1}^K {\left\| {A_n } \right\|} $.

Now lets prove for $\displaystyle K+1$, so say $\displaystyle \{A_n:n=1,2,\cdots,K+1\}$ is a collection of pairwise disjoint finite sets [/tex].

Observe that $\displaystyle \bigcup\limits_{n = 1}^{K + 1} {A_n } $ is just $\displaystyle \left( {\bigcup\limits_{n = 1}^K {A_n } } \right) \cup A_{K + 1} $

Let $\displaystyle \mathbf{C}=\left( {\bigcup\limits_{n = 1}^K {A_n } } \right) $.

Note that $\displaystyle \mathbf{C} \cap A_{K+1}=\emptyset$.

Apply the basic Counting Principle. - Mar 24th 2011, 02:24 PMqwerty10
To prove the basic counting principle do we just simply state AuB=A + B - AnB and AnB=empty set so we just get AuB= A + B? I know there should be cardinality signs around the sets

- Mar 24th 2011, 03:36 PMPlato
Is that a statement? Or is it a question?

Why not learn to post in symbols? You can use LaTeX tags

[tex]\|A\|[/tex] gives $\displaystyle \|A\| $