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I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

I take from $a$ some of its elements, say integers $10$, $11$, and $12$ only.

So now we have a new set $a'$ that is the infinite set of all nonnegative integers $0,1,2,3...$ except for $10$, $11$, and $12$.

If I subtract: $a-a'$ the result is a set comprised of $10$, $11$, $12$ only. Is this correct? Can one subtract infinities like this?

If yes, does this mean that $a \gt a'$(despite that both are infinite)?

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$0$ is not positive and "set a > new set a" does not make sense -- what are you trying to compare here? –  oldrinb May 11 '13 at 21:35
    
let's say the sum of the set to make it simple –  Ray S. May 12 '13 at 5:32

2 Answers 2

up vote 5 down vote accepted

The set of positive integers is denoted $\mathbb{Z}^+=\{1,2,3,\dots\}$ and has cardinality $\aleph_0$. The set of positive integers excluding $10,11,12$ also has cardinality $\aleph_0$. The difference between the sets is indeed $\{10,11,12\}$ and its cardinality is $3$. To answer your question, you should read more about cardinal number arithmetic.

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Subtraction between infinite cardinals cannot be well-defined. This is a good example why.

We know that $|\Bbb N|$ and $|\Bbb Z|$ are both of the same cardinality, but so is $|\Bbb{Z\setminus N}|$. On the other hand, $|\Bbb N\setminus\{k\in\Bbb N\mid k>2\}|=3$ (zero is a natural number here).

So we have $\aleph_0-\aleph_0=\aleph_0$ but at the same time $\aleph_0-\aleph_0=3$ and we can easily engineer this to be any other result between $0$ and $\aleph_0$ as well. This is indeterminate much like $\infty-\infty$ is an indeterminate form in calculus, although $\infty$ and $\aleph_0$ are not directly related (these are two different, and mostly incompatible, notions of infinity).

The reason is that for infinite sets being a proper subset does not imply having strictly smaller cardinality like it does with finite sets. Our intuition from finite mathematics usually does not apply to infinite objects.

On the other hand, subtracting two concrete sets is perfectly doable and legitimate mathematically. But we cannot conclude anything on the cardinality of the result, unless we know what are the specific sets (or at least more than just "two sets", which is the most general notion).

Read more about this:

  1. Why do the rationals, integers and naturals all have the same cardinality?
  2. Is there a way to define the "size" of an infinite set that takes into account "intuitive" differences between sets?
  3. Why the principle of counting does not match with our common sense
  4. Cardinality of subtraction of sets
  5. Cardinal number subtraction (See my answer, as it is related to the comment below by user14111.)
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I'm still semi-sure that there is a duplicate hiding underneath all that website, but I couldn't find it. I did find six similar, related, but not duplicates. So instead I posted this answer (with four of the links I found, too). –  Asaf Karagila May 11 '13 at 21:55
    
@user14111: If it's not always well-defined then it's not well-defined. Even division by zero is defined if we consider $\frac00=0$, but $\frac30$ is not defined anymore. (The former comes from the definition in $\Bbb N$ as $k\mid n\iff\exists m.(km=n)$ and the least such $m$ is defined as $\frac nk=m$ in that case. –  Asaf Karagila May 11 '13 at 22:38
    
@user14111: Unless your context explicitly states that a well-defined function doesn't have to be well-defined on the entire domain, yes. The common notion is that being well-defined means that you are well-defined. Let me ask you a very natural question, then what is $2^{\aleph_0}-\aleph_1$? Is it defined at all? If not, what is $2^{\aleph_1}-2^{\aleph_0}$? Is it defined at all? And I can continue in this fashion infinitely many times. So while sometimes seeking maximal domain is a good idea, in this context we have too many undecidable questions to make such decision reasonable. –  Asaf Karagila May 12 '13 at 1:58
    
@user14111: As for the point about AC, do note that it actually becomes much more interesting when assuming that the axiom of choice fails acutely. Dedekind-finite sets do have the property that the subtraction of two Dedekind-finite cardinals is well-defined. One can have a proper class of those cardinals, and get interesting results too. –  Asaf Karagila May 12 '13 at 2:00

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