"It is impossible to define addition of cardinalities since the resulting operation is not well-defined"

The above is the true and false question and what i think the statement above is false and my reasoning is below by giving counter example:

$$|A|+|B|=|\{(a,\#)\mid a\in A\}\cup\{(b,*)\mid b\in B\}\;|.$$ is well defined.

Can anyone correct me!!!

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I edited your question a bit to reflect what I think you meant to ask. Please have a look to see if it agrees with what you wanted to ask. – Ittay Weiss Feb 14 '13 at 3:28
@ittayweiss the statement mark with inverted comas is what im trying to prove whether it is true or false... – Timoci Lagilevu Feb 14 '13 at 3:32
notice then that on the right hand side of the equality you have a set while on the left hand side a sum of cardinalities. You probably meant to put the cardinality symbols around the set on the right hand side too. – Ittay Weiss Feb 14 '13 at 3:36
ohh yes i forgot thankx @ittayweiss – Timoci Lagilevu Feb 14 '13 at 3:39

The operation is well-defined, provided that $\#$ and $*$ are understood to be distinct symbols. The point is that

$$|A|=\big|\{\langle a,\#\rangle:a\in A\}\big|\;,\tag{1}$$

because the map $a\mapsto\langle a,\#\rangle$ is a bijection, and similarly

$$|B|=\big|\{\langle b,*\rangle:b\in B\}\big|\;,\tag{2}$$

and (very important!)

$$\{\langle a,\#\rangle:a\in A\}\cap\{\langle b,*\rangle:b\in B\}=\varnothing\;.$$

Because the sets $\{\langle a,\#\rangle:a\in A\}$ and $\{\langle b,*\rangle:b\in B\}$ are disjoint, the cardinality of their union is just the sum of their cardinalities, which by $(1)$ and $(2)$ is $|A|+|B|$ as we normally understand it.

That is, even if $A$ and $B$ overlap, so that $|A\cup B|\ne|A|+|B|$, the sets $\{\langle a,\#\rangle:a\in A\}$ and $\{\langle b,*\rangle:b\in B\}$ do not overlap, and their union therefore does have the desired cardinality.

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you mean the statement marked with the inverted commas in the question above is true? – Timoci Lagilevu Feb 14 '13 at 3:45
@Timoci: I don’t understand: I explicitly said that it is false, and that the operation as defined here is well-defined. – Brian M. Scott Feb 14 '13 at 3:46

If we want to claim that the definition $$|A|+|B|=|(A\times\{\#\})\cup (B\times\{*\})|$$ is well-defined, then we actually want to say the following:

Let $|A|=|C|$ and $|B|=|D|$ then $|A|+|B|=|C|+|D|$.

Suppose $f\colon A\to C$ is a bijection and $g\colon B\to D$ is a bijection, these bijections exist because we assumed $|A|=|C|$ and $|B|=|D|$, then the following is a $h$ bijection witnessing $|A|+|B|=|C|+|D|$: $$h(\langle x,y\rangle) = \begin{cases} \langle f(x),y\rangle & y=\#\\\langle g(x),y\rangle & y=*\end{cases}$$

Namely given a pair, if it came from $A\times\{\#\}$ then we apply $f$ to the left coordinate, and we have a pair from $C\times\{\#\}$; and if $y=*$ then the pair was coming from $B\times\{*\}$ and then we apply $g$ to the left coordinate and we have a pair from $D\times\{*\}$.

I leave it to you to verify that indeed $h$ is a function, its domain is $(A\times\{\#\})\cup(B\times\{*\})$, and its range is $(C\times\{\#\})\cup(D\times\{*\})$, and that it is indeed a bijection.

Remember that these unions are disjoint unions!

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