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I've been working on a problem and when I looked around the site I couldn't seem to find much that looked like what I was trying to solve. My problem is as follows:

Let $A$ be an infinite set and denote card($A$) by $\alpha$. If $B$ is an infinite set, denote card($B$) by $\beta$. Define $\alpha\beta$ to be card($A\times B$). Let $B'$ be a set of disjoint from $A$ such that card($B$)=card($B'$). Define $\alpha+\beta$ to be card ($A\cup B'$). Denote by $B^A$ the set of all maps of $A$ into $B$, and denote card($B^A$) by $\beta^{\alpha}$. Let $C$ be an infinite set and abbreviate card($C$) by $\gamma$.

Prove (a). $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$.

Attempt: Let $f:A\times(B\cup C')\rightarrow(A\times B)\cup(A\times C')$ be defined by $f(a,x)=(a,x)$ where card$(C')=$card$(C)$ and $C'$ is disjoint from $B$. Suppose that $f(a,x)=f(b,y)$. Then $(a,x)=(b,y)$ so $f$ is injective. Now, if $(a,x)\in(A\times B)\cup(A\times C)$ then $f(a,x)=(a,x)$ so $f$ is surjective. Therefore, $f$ is bijective and the claim is true.

My thought process is that $A\times(B\cup C')=(A\times B)\cup(A\times C')$ so the identity map is a bijection which gives me that the cardinalities are the same.

(b) $\alpha\beta=\beta\alpha$

Attempt: Let $f:A\times B\rightarrow B\times A$ be defined by $f(a,b)=(b,a)$. Now, suppose that $f(a,b)=f(c,d)$. Then $(b,a)=(d,c)$ so $(a,b)=(c,d)$ and we conclude that $f$ is injective. Now, if $(b,a)\in B\times A$ then $f(a,b)=(b,a)$ so $f$ is surjective. Therefore, $f$ is bijective so the claim is true.

This one seems fairly clear to me although I will gladly accept any criticism.

(c) $\alpha^{\beta+\gamma}=\alpha^{\beta}\alpha^{\gamma}$.

Attempt: Let $\phi:A^{B\cup C'}\rightarrow A^B\times A^{C'}$ be defined by $\phi(\alpha)=(\beta,\delta)$ where $\alpha:B\cup C'\rightarrow A$, $\beta:B\rightarrow A$, and $\delta:C'\rightarrow A$. Furthermore, $(\beta,\delta)(x)=\beta(x)$ if $x\in B$ or $\delta(x)$ if $x\in C'$. Suppose that $\phi(\alpha)(x)=\phi(\sigma)(x)$ where $\phi(\sigma)=(\rho,\tau)$. Then $(\beta,\delta)(x)=(\rho,\tau)(x)$ so if $x\in B$ then $\beta(x)=\rho(x)$ otherwise if $x\in C'$ then $\delta(x)=\tau(x)$. Thus, $\beta=\rho$ and $\delta=\tau$ so $\alpha=\sigma$ and we conclude that $\phi$ is injective. Now, if $(\beta,\delta)\in A^B\times A^{C'}$ then $\phi(\alpha)(x)=(\beta,delta)(x)$ so $\phi$ is surjective. Therefore, the claim is true.

I admit I am a bit lost on this one. I've tried to define a function using the given definitions but am confused on having a product of two sets of maps and how they relate.

Any help would be appreciated.

Andrés E. Caicedo
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JohnC
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  • Surely $A \times (B \cup C') = (A \times B) \cup (A \times C')$? – copper.hat Nov 09 '17 at 03:56
  • For (c), let $\phi(\alpha) = (\alpha \mid_B, \alpha \mid_{C'})$. Since $B,C'$ are disjoint, it is easy to define and validate an inverse. – copper.hat Nov 09 '17 at 04:04
  • Why do you have quotes in the title? – Asaf Karagila Nov 09 '17 at 04:13
  • https://math.stackexchange.com/questions/131212/overview-of-basic-results-on-cardinal-arithmetic and https://math.stackexchange.com/questions/57398/how-to-show-abc-abc-for-arbitrary-cardinal-numbers by the way. – Asaf Karagila Nov 09 '17 at 04:14
  • My point is that cardinal arithmetic is an actual thing. It is not "arithmetic" just because it is different, anymore than a chimpanzee a "living thing" because it differs from the living things usually found around me. The links are there because they contain many relevant threads that might help you. I didn't close this as a duplicate, because this is also a proof verification question. – Asaf Karagila Nov 09 '17 at 14:02
  • @copper.hat I'm having issue wrapping my head around your definition for $\phi$. If $x\in B$ then $\phi(\alpha(x))=(\alpha(x),$undefined)? I'm not sure how to assign a value for $\alpha_{C'}(x)$ where $x\not\in C'$. Would it work to set $\alpha(x)=0$ if $x$ is not in the domain of the map? – JohnC Nov 09 '17 at 21:35
  • @JohnC: I added an elaboration as a partial answer. – copper.hat Nov 10 '17 at 19:26

2 Answers2

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My thought process is that $A\times(B\cup C')=(A\times B)\cup(A\times C')$ so the identity map is a bijection which gives me that the cardinalities are the same.

$\def\card{\operatorname{card}}$Yes, you have that those sets are identical, so then just use the definitions you have been given to show the cardinality expressions are equal. $$X=Y \implies \card (X)=\card(Y)$$

Declair that $A',B',C'$ to be pairwise disjoint sets with the same cardinalities as $A, B, C$ respectively, namely, $\alpha,\beta,\gamma$.

We have by the definitions: $~\card(A'\times(B'\cup C')) ~{=\card(A')\card(B'\cup C') \\= \card(A')(\card(B')+\card(C')) \\= \alpha(\beta+\gamma)}$

We likewise have : $~\card((A'\times B')\cup(A'\times C')) ~{=\card(A'\times B')+\card(A'\times C') \\= \card(A')\card(B')+\card(A')\card(C') \\= \alpha\beta+\alpha\gamma}$

Well, since the sets are the same, the cardinalities shall be equal, so: $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$.

$\Box$

Which seems trivial, except that the cardinalities are not real numbers.$

Graham Kemp
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Not a complete answer, just too long for a comment.

When I write $\phi(\alpha) = (\alpha \mid_B, \alpha \mid_{C'})$, I mean that $\alpha : B \cup C' \to A$ and $\phi(\alpha)$ is the tuple of functions $\alpha \mid_B : B \to A$ and $\alpha \mid_{C'} : B \cup C' \to A$. Then we have $\alpha \mid_B(x) = \alpha(x)$ for $x \in B$ and $\alpha \mid_{C'}(x) = \alpha(x)$ for $x \in C'$.

The inverse is defined as $\phi^{-1} (\beta , \gamma) (x) = \begin{cases} \beta(x), & x \in B \\ \gamma(x), & x \in C' \end{cases}$.

It follows that $\phi:A^{B\cup C'}\rightarrow A^B\times A^{C'}$ is bijective.

copper.hat
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