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In reading Kevin Houston's "How to Think Like a Mathematician", there's a line stating the following:

Let $X$ be the set of finite sets. Then the cardinality of a set is a function on $X$, that is $|.|: X \to \{0\} \cup \mathbb{N}$. Note that we need $0$ in the co-domain as the set could be the empty set.

What does this mean exactly? From my understanding, this would imply that any finite set is mapped exactly to one natural number. For example, a finite set of $5$ elements would be mapped to the natural number $5$. However, is it possible to map the set itself to a value?

quid
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Anya
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3 Answers3

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It means what you think it means.

The domain of definition for this function is the "set of all finite sets".

Thus every element here is a "finite set".

Consider for a simpler example: $A=\{\{1,2\}, \{1,2,3\},\{5,7,8,12\}\}$

This is a set whose elements are the sets $\{1,2\}$ and $\{1,2,3\}$ and $\{5,7,8,12\}$.

If you define a function on $A$ then you assign a value to each of $\{1,2\}$ and $\{1,2,3\}$ and $\{5,7,8,12\}$ yet not to $1,2,3, 5,7,8,12$.

Tangential remark: there is a problem with the idea of a set of all finite sets, but this is not relevant in this context.

quid
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  • But then how would you infer the cardinality of the set from that? What "number" is assigned to each of {$1,2$} etc.? – Anya Oct 15 '18 at 17:13
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    The function is defined as assigning to ${1,2}$ the number of elements in it, that is $2$. And for ${5,7,8,12}$ it would be $4$. To ask for the number of elements in the number $13$ does not make sense, thus the function you consider is defined for sets only. – quid Oct 15 '18 at 17:18
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You understood wrong. According to this definition, any finite set is mapped exactly to a set whose cardinality is a natural number or zero.

  • That seems incorrect to me. – quid Oct 15 '18 at 17:04
  • The above comment or my understanding? If that's true, then what does the above mentioned statement imply about the elements in the set that $X$ is being mapped to? – Anya Oct 15 '18 at 17:07
  • @quid Every cardinal number is a set. – Jakobian Oct 15 '18 at 17:10
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    @Jakobian sure, and the set of all finite sets does not even exist. Please, don't confuse the poster. In any case the definition in OP has as image the natural numbers. (And of course we can consider a natural number as a set, but that's besides the point here.) – quid Oct 15 '18 at 17:12
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For every finite set, its function value is its cardinality. So $\{1,2,3\}$ is mapped to $3$ just as $\{2,4,8\}$ is, as both have the same cardinality $3$. $\emptyset$ is mapped to $0$ (as the only set with that property).

I don't quite see what the problem is: there is a class of all finite sets and we can define a function with those sets as input.

Henno Brandsma
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  • You don't see the problem because you probably don't remember not knowing this. Anya is a beginner who was taught that functions map elements from one set to another. Anya has probably never considered how complex those sets can be. – John Douma Oct 15 '18 at 17:21
  • @JohnDouma a set of sets is perfectly valid. I was taught that at age 13 in high school.. – Henno Brandsma Oct 15 '18 at 17:22
  • That's my point. You have known this for a long time and this is easy for you. It is easy for me too, but not for everyone. – John Douma Oct 15 '18 at 17:35