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I define equinumerous and cardinality in this way:

$A$ and $B$ are equinumerous (written $A\sim B$) if there is a bijection between them.

We say $card(X)=card(Y)$ if $X\sim Y$.

I would like to prove the cardinality is well-defined. This is how I did (I'm using the symmetry and transitive properties):

$X\sim Y$ and $X\sim Z\implies Y\sim Z\implies card(Y)=card(Z)$.

Am I correct? Do I need to check if the concept of equinumerous is well-defined as well?

Thanks

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  • $\begingroup$ What do you mean by welldefined? Do wish to show that it is an equivalence relation? $\endgroup$
    – user45150
    Apr 4 '15 at 17:12
  • $\begingroup$ @user45150 I want to show these two notions are not ambiguous. $\endgroup$
    – user42912
    Apr 4 '15 at 17:13
  • $\begingroup$ @user45150 I know how to prove this is an equivalence relation. $\endgroup$
    – user42912
    Apr 4 '15 at 17:14
  • $\begingroup$ What do you mean by non ambiguous notions? It's ambiguous for me! $\endgroup$
    – user226387
    Apr 4 '15 at 17:16
  • $\begingroup$ I guess I am confused. $\text{card}(X)=\text{card}(Y)$ if and only if there exists a bijection between $X$ and $Y$. This is a legitimate definition. You don't pick a representative for an equivalence set or any other of the usual well-definedness problems. $\endgroup$
    – user45150
    Apr 4 '15 at 17:16
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You could also think that when you say "$card(X)=card(Y)$ iff $X\sim Y$", you're meaning:

"Assign every set $X$ a number called $card(X)$ in such a way that $card(X)=card(Y)$ iff $X\sim Y$"

In fact, you need to prove that $\sim$ is an equivalence relation in order to have a good definition of this number. Later on, you define $card(X)$ for known sets as $\{1,...,n\}$ or $\mathcal{P}(\mathbb{N})$, for example.

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  • $\begingroup$ So using this definition we don't need to prove the well-definition of the finite cardinality, i.e., if there are bijections $f:I_n\to X$ and $g:\to I_m$, then $m=n$ (notation: $I_n=\{1,\ldots,n\}$) $\endgroup$
    – user42912
    Apr 4 '15 at 17:28
  • $\begingroup$ No. You still need to prove it. In fact, this definition lets you define the cardinality of $I_n$ as $m$ (since you'd have $I_n\sim I_m$), but, on the other hand, you define the cardinality of $I_n$ as $n$, so you must prove those are equal. (PS: You missed an $X$ after the $g$). $\endgroup$
    – Daniel
    Apr 4 '15 at 17:34
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    $\begingroup$ @DanielEscudero The discussion you give above may be a little confusing, because you do not mention what $\text{card}(X)$ actually is. For instance what happens if $X$ is infinite. I agree that the conflation of the definition of cardinality that you give and the one in the original question may be the cause of the confusion. The reason I did not mention this point of view in my comments was that it was not mentioned in the question and I did not want to add confusion, but since it is brought up we may as well develop it. $\endgroup$
    – user45150
    Apr 6 '15 at 15:54
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    $\begingroup$ @user42912 I think what is important to remember with regards to Daniel Escudero's definition is that it is a way to associate a size, called the cardinality, to every set $X$. For this you must check that $\sim$ is an equivalence relation because if you associate the number $n$ to $X$ and $m$ to $Y$ and $X\sim Y$ you would like to be able to conclude that $n=m$. $\endgroup$
    – user45150
    Apr 6 '15 at 15:58
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    $\begingroup$ @DanielEscudero Another thing I forgot to mention about your response is that it should probably be checked that $I_n\not\sim I_m$ if $n\neq m$. That is implicit in what your method needs for well-definedess (but not what was needed in user45150's original question) $\endgroup$
    – user45150
    Apr 6 '15 at 16:00

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