• Mar 24th 2011, 01:03 PM
qwerty10
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 PM
Plato
Quote:

Originally Posted by qwerty10
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)

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 PM
qwerty10
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 PM
Plato
Quote:

Originally Posted by qwerty10
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

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 PM
qwerty10
The theorem is to do with counting so that must be the notation
• Mar 24th 2011, 01:56 PM
Plato
Quote:

Originally Posted by qwerty10
The theorem is to do with counting so that must be the notation

There is a basic counting principle: $\displaystyle \|A\cup B\|=\|A\|+\|B\|-\|A\cap B\|$.
Now for disjoint sets that last term is just zero.
In other words for disjoint sets $\displaystyle \|A\cup B\|=\|A\|+\|B\|$.
To prove your theorem use induction.
• Mar 24th 2011, 02:01 PM
qwerty10
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 PM
Plato
Quote:

Originally Posted by qwerty10
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

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 PM
qwerty10
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 PM
Plato
Quote:

Originally Posted by qwerty10
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

Is that a statement? Or is it a question?

Why not learn to post in symbols? You can use LaTeX tags
$$\|A\|$$ gives $\displaystyle \|A\|$