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Math 3345 Section 16 Exercise 9

Let A, B, C, D be sets such that A is equinumerous to C and B is equinumerous to D. Show that AxB is equinumerous to CxD.

I believe I have the correct answer for when all A, B, C, D are finite. I think it works but I'm not sure if it's correct for when the sets are infinite.

My answer: I'm using the '||' symbols to denote "the cardinality of"/"the length of", and 'x' to represent Cartesian product (I apologize, I'm new to formatting on the site)

A equinumerous to C, therefore |A| = |C|

B equinumerous to D, therefore |B| = |D|

|AxB| = (|A|)(|B|) = (|C|)(|B|) = (|C|)(|D|) = |CxD|

Therefore, since |AxB| = |CxD|, then AxB is equinumerous to CxD.

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  • $\begingroup$ Why can you do $|A\times B|=|A||B|$? $\endgroup$ – YTS Nov 30 '14 at 19:08
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    $\begingroup$ That is, essentially, what you have to prove. That this calculation is valid. $\endgroup$ – Asaf Karagila Nov 30 '14 at 19:09
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    $\begingroup$ As @Asaf implies, your argument is essentially circular. You need to get back to fundamentals. By hypothesis you have bijections $f:A\to C$ and $g:B\to D$. How can you use $f$ and $g$ to construct a bijection between $A\times B$ and $C\times D$? $\endgroup$ – Brian M. Scott Nov 30 '14 at 19:11
  • $\begingroup$ It's just rule we introduced earlier in the semester by using the multiplication rule: for two finite sets A, B the cardinality of A x B = cardinality of A times cardinality of B. $\endgroup$ – Draggfire654 Nov 30 '14 at 19:12
  • $\begingroup$ But you can not argument in the same way in the infinite case. See my answer below. $\endgroup$ – YTS Nov 30 '14 at 19:21
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As Asa Karagila said your argument is not right since essentially is what you have to prove (at least in the infinite case)

Try the following:

As $|A|=|C|$ then there exists a bijection $f:A\rightarrow C$. Analogous there exist a bijection $g:B \rightarrow D$.

The function $f\times g :A\times B\rightarrow C\times D$ defined as $(f\times g) (x,y)=(f(x),g(y))$ is a bijection. Prove it.

Finally conclude $|A\times B|=|C\times D|$.

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  • $\begingroup$ I suppose I don't really know specifically how. I understand what a bijection is, and I know in order to prove a bijection exists you must prove that a function is both injective and surjective. But how to show that in specific cases I do not understand. $\endgroup$ – Draggfire654 Nov 30 '14 at 19:36
  • $\begingroup$ Inyective: $f(x)=f(y)$ implies $y=x$. Then you hace to pick two elements with the same image and conclude that the elements are equal Surjetive: there exist for each $y$ in the range an element $x$ in the domain such that $f(x)=y$. Then you have to pick one element in the range and find an element in the domain such that it is its image. $\endgroup$ – YTS Nov 30 '14 at 19:45

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