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I'm trying to prove that for any cardinal numbers $a,b,c$, the following holds: $a ^ {b + c} = a ^ b a ^ c$ i.e. that there exists a bijective function $ f : A ^ {B \:\: \cup \:\: C} \rightarrow A^B \times A^C $

This is only part of the proof sketch I have (proving $f$ is injective), and I'd like to know if is well written, since I believe it has flaws.


Let $f: \{ g \:\:\: | g:B \cup C \rightarrow A \} \rightarrow \{ \langle g,h\rangle | \:\:\: g: B \rightarrow A \wedge h : C \rightarrow A \}$ such that

$f ( g_{b} \cup g_{c}) = \langle g_{b},g_{c}\rangle$.

Now,

$f( g_{b1} \cup g_{c1}) = f ( g_{b2} \cup g_{c2} ) \implies \langle g_{b1},g_{c1}\rangle = \langle g_{b2},g_{c2}\rangle$ and therefore $f$ is injective.


Questions:

  1. Does that prove that $f$ is injective? I think it does not, since $f ( g_{b} \cup g_{c} ) = f ( g_{c} \cup g_{b}) \implies \langle g_{b},g_{c}\rangle = \langle g_{c},g_{b}\rangle (\bot)$
  2. Is there an alternative way to define $f$? It's difficult for me to define it in terms of properties of elements of its domain.

Side note:

The title says "kinds" because the function domain and image sets are sets of sets, but I may be mistaken using that word, if so, please edit accordingly.

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In ZF(C), the most common axiomatization of set theory, all sets are sets of sets, so I’ve just put a description of $f$ in the title. –  Brian M. Scott Nov 8 '11 at 5:23
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What do you mean by $g_b\cup g_c$? Are you trying to say that $f:g\mapsto\langle g\upharpoonright B,g\upharpoonright C\rangle$? That function is indeed injective. It won’t be surjective, however, unless $B\cap C=\varnothing$. –  Brian M. Scott Nov 8 '11 at 5:25
    
Rather than $g_{b}\cup g_{c}$, I would write simply $g$, and map it to $\langle g|_B,g|_C\rangle$, where $g|_B$ is the restriction of $g$ to $B$, and $g|_C$ the restriction to $C$. To prove injectivity, your notation is not very nice, so instead you should us $f(g) = f(h)$ and note that this yields $g|_B=h|_B$ and $g|_C = h|_C$. Etc. P.S. Use \langle and \rangle for $\langle$ and $\rangle$ rather than < and >; the spacing is off for the latter. –  Arturo Magidin Nov 8 '11 at 5:27
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Is the union in $A^{B\cup C}$ meant to be disjoint union? Otherwise, the desired bijection need not exist. –  Arturo Magidin Nov 8 '11 at 5:28
    
For the argument to work as stated, we probably need to ask that $B$ and $C$ are disjoint. Why not define $\phi(g)$ as the ordered pair $(g_B,g_C)$? A lot fewer symbols. (The $\phi$ is because somehow $f$ sounds like something of the same family as $g$.) The injectivity should be easy. From $(g_B, g_C)$ we can reconstruct $g$. –  André Nicolas Nov 8 '11 at 5:34
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3 Answers 3

up vote 2 down vote accepted

You can think of $g : B \cup C \to A$ as a couple of functions $g_B : B \to A$ and $g_C : C \to A$ such that for all $x \in B \cap C$, $g_B(x) = g_C(x)$. Therefore, letting $f : \Gamma \to \Delta$ (with $$ \Gamma = \{ g \,\, | \,\, g : B \cup C \to A \}, \qquad \Delta = \{ \langle g_B, g_C\rangle \, \, | \, \, g_B : B \to A, \, \, g_C : C \to A \} $$ obviously), you can see that a natural way to do this is to restrict the domain of $g : B \cup C \to A$ to just $B$ (or $C$), hence the map $f$ could naturally be defined by $g_B(x) = g(x)$ for all $x \in B$, $g_C(x) = g(x)$ for all $x \in C$ and $f(g) = \langle g_B, g_C \rangle$. In this manner the notation is more clear and more obviously well-defined.

This function $f$ is injective because of the following. Suppose that $f(g_1) = \langle g_B^1, g_C^1 \rangle = \langle g_B^2, g_C^2 \rangle =f(g_2)$. Thus for all $x \in B \cup C$, $x$ is either in $B$ or $C$, suppose $B$. Thus $g_1(x) = g_B^1(x) = g_B^2(x) = g_2(x)$. The case $x \in C$ is similar. Therefore $g_1 = g_2$.

I can't think of a natural way to define $ F : \Delta \to \Gamma$ right now, it'll require some thinking, I guess. I am not familiar with the usual rigor abusive context, but my problem is that I don't want to assume $B \cap C = \varnothing$. If we can assume that there is an obvious way back (i.e. you can show that this $f$ is bijective, easily). If someone could comment on this question I'm asking, I'd love to read.

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There’s not much reason to define an $F:\Delta\to\Gamma$ unless $B\cap C=\varnothing$, in which case the desired $F$ is simply $F:\Delta\to\Gamma:\langle g,h\rangle\mapsto g\cup h$. –  Brian M. Scott Nov 8 '11 at 5:31
    
Yeah... I know it would sound lame to do it if the intersection was non-empty, but ZFC theory usually scares me by its depth and generality, so I don't know if it is allowed sometimes in some contexts. I've never actually studied ZFC rigorously so I preferred to ask instead of babbling. –  Patrick Da Silva Nov 8 '11 at 5:32
    
Perhaps because of this my notation seems more familiar to a mathematician that doesn't study logic or set theory because I've understood but never seen the notation $g \cup h$ for taking the "union" of two functions... –  Patrick Da Silva Nov 8 '11 at 5:34
    
The functions $g$ and $h$ are just sets of ordered pairs, so their union is also a set of ordered pairs. As long as $g$ and $h$ don’t contain two pairs $\langle b,a\rangle$ and $\langle c,a\rangle$ with $b\ne c$, their union will still be a function $-$ and if $B\cap C=\varnothing$, they won’t contain such conflicting pairs. –  Brian M. Scott Nov 8 '11 at 5:49
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Thank you Brian, Pattrick and yunome for your help, I really appreciate it. –  kmels Nov 8 '11 at 6:19
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I’m going to assume that I’ve guessed correctly that you want to define $$f:A^{B\;\cup\;C}\to A^B\times A^C:g\mapsto \langle g\upharpoonright B,g\upharpoonright C\rangle\;.$$ If you express it this way, it’s easy to see that $f$ is injective. Suppose that $f(g_0)=f(g_1)$. Then $g_0\upharpoonright B=g_1\upharpoonright B$ and $g_0\upharpoonright C=g_1\upharpoonright C$. Now let $x\in B\cup C$ be arbitrary. If $x\in B$, then $$g_0(x)=(g_0\upharpoonright B)(x)=(g_1\upharpoonright B)(x)=g_1(x)\;,$$ and if $x\in C$, then $$g_0(x)=(g_0\upharpoonright C)(x)=(g_1\upharpoonright C)(x)=g_1(x)\;,$$ so $g_0(x)=g_1(x)$ for all $x\in B\cup C$, and hence $g_0=g_1$. It was the clumsy way in which you expressed an arbitrary element of $A^{B\;\cup\;C}$ that led you to doubt the injectivity of $f$.

Note that although this $f$ is injective, it is not surjective if $B\cap C\ne\varnothing$ and $A$ has at least two elements. To see this, suppose that $B\cap C\ne \varnothing$, and fix $x\in B\cap C$. Let $g_0\in A^B$ be arbitrary, and let $g_1\in A^C$ be any function such that $g_1(x)\ne g_0(x)$. Then $\langle g_0,g_1\rangle \ne f(h)$ for any $h\in A^{B\;\cup \;C}$.

To get bijectivity, you must assume that $B\cap C=\varnothing$. In that case $f$ as defined above is a bijection, and $$f^{-1}:A^B\times A^C\to A^{B\;\cup\;C}:\langle g,h\rangle\mapsto g\cup h\;.$$

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Thanks for pasting the exact same argument as mine but adding $B \cap C = \varnothing$... you could've just ask me to complete my answer instead of making a new one though. (I am not arguing, I am thanking.) –  Patrick Da Silva Nov 8 '11 at 6:03
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I don't believe this proves $f$ is injective, for if $f(g_{b1}\cup g_{c1})=\langle g_{b1},g_{c1}\rangle$, then $f( g_{b1} \cup g_{c1}) = f ( g_{b2} \cup g_{c2} )$ and$\langle g_{b1},g_{c1}\rangle = \langle g_{b2},g_{c2}\rangle$ are really just the same statement so you haven't really proven anything.

I assume you also have the assumption that $B$ and $C$ are disjoint, for if not, surjectivity does not necessarily follow, and you want to show that $f$ is a bijection. I think the best way to go about proving injectivity is to first define $f$ by $f(g)=\langle g|_B,g|_C\rangle$, where $g|_B$ is the restriction of $g\colon B\cup C\to A$ to $B$, and likewise for $C$.

There are are two usual ways to show injectivity here. One is to show that $f(x)=f(y)\implies x=y$ or its contrapositive $x\neq y\implies f(x)\neq f(y)$.

One way to use it here is to suppose $g\neq h$ for $g,h\in A^{B\cup C}$, and then show $f(g)\neq f(h)$. Then for some $x\in B\cup C$, $g(x)\neq h(x)$. Let's also assume $x\in B$. Then $f(g)=\langle g|_B,g|_C\rangle$ and $f(h)=\langle h|_B,h|_C\rangle$. However, $g|_B\neq h|_B$ since $g(x)\neq h(x)$ for $x\in B$, so $f(g)\neq f(h)$, and $f$ is injective. The same idea works if you had chosen $x\in C$.

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