Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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!!!

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer
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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.