# Proving that the notion of difference of cardinalities is not well-defined.

Consider the following operation on cardinalities. Given sets A,B write |A|-|B|=|A-B|.

Prove that this notion of difference of cardinalities is not well-defined.

Proof:

To Prove that |A|-|B|=|A-B| is not well-defined we will give counter example.

To begin with;

Let A={a,b,c,d,e} and B={h,i,j}

From the above, it is clearly seen that |A|=5 and |B|=3

If we consider the Left Hand Side:

|A|-|B|=5

Now consider the Right Hand Side:

|A-B|=2

Since there is no bijection between |A|-|B| and |A-B|, it is then concluded that the notion of difference of cardinalities is not well-defined.

Can anyone correct me on this please!!!

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The idea’s fine, but you have them backwards: $|A|-|B|=5-3=2$, and $|A\setminus B|=|A|=5$. – Brian M. Scott Feb 7 '13 at 1:21
thankx @Brian M. Scott – Timoci Lagilevu Feb 7 '13 at 1:37
@BrianM.Scott I must disagree here. – Ittay Weiss Feb 7 '13 at 1:41
@Ittay: It’s fine as far as it goes; it just doesn’t go far enough. I was urgently called away and left the comment unfinished. I’d intended to add the necessity of looking at a $B$ like $\{a,b,c\}$, say, but by the time I got back, you’d answered, so I didn’t bother. – Brian M. Scott Feb 7 '13 at 1:47
I figured something of the sort, but just did not want the unfinished comment to throw the unsuspecting off... – Ittay Weiss Feb 7 '13 at 1:54

To show that the notion given is not well defined you have to do the following: Find four sets $A,B,C,D$ such that $|A|=|B|$ and $|C|=|D|$ while $|A|-|C|\ne |B|-|D|$. In your answer above you did not do that. Moreover, your claim that "there is no bijection between $|A|-|B|$ and $|A-B|$" is wrong since the notion of difference you are considering says explicitly that $|A|-|B|$ is $|A-B|$.

You need to work with the notion of equality of cardinalities whereby $|X|=|Y|$ precisely when there exists a bijection $X\to Y$, and not rely on your existing intuition regarding counting finite sets. Again, you can present your counter example by exhibiting four sets as I explained above.

EDIT: To clarify what I mean then: A source of confusion about such questions is to think that one shows the notion not to be well-defined if one shows that notion does not agree with ordinary counting. That is a wrong conclusion though and if one sticks to the precise definitions of cardinalities instead of reverting to counting then one avoids this pitfall. The reason the notion is ill-defined is not it's weird computational results but their inconsistency. For instance, defining $|A|-|B|=|\emptyset |$ is a well-defined notion. It gives weird results that to us have nothing to do with subtraction but it is a well-defined notion. The notion in the question though is not well-defined. Totally different.

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When you say not to "rely on your existing intuition regarding counting finite sets" this suggests that you think some of $A,B,C,D$ must be infinite. This isn't true with the OP's definition of subtraction: you can show it's ill-defined using finite sets. – Trevor Wilson Feb 7 '13 at 1:42
...for example, if $C$ and $D$ are nonempty and $C$ is a subset of $A$ but $D$ is not a subset of $B$. – Trevor Wilson Feb 7 '13 at 1:44
You are absolutely right @TrevorWilson, and I did not mean to produce a counter example with infinite set. I added a bit more text to, I hope, explain my comment. – Ittay Weiss Feb 7 '13 at 1:48
Ok, now it is clear---the definition of subtraction is bogus, so we can't expect it to be well-defined the way we would expect the better definition (the one where we set set being subtracted is a subset) to be. At first I thought you were alluding to the fact that even the better definition is ill-defined for infinite sets. – Trevor Wilson Feb 7 '13 at 1:51