Set equality

**kuntah** · January 24th 2008, 02:51 PM

I need to prove this plz help

$ A,B,C \ and\ D\ are\ non\ empty\ sets\ with\ \left|A\right|=\left|B\right| and\ \left|C\right|=\left|D\right|$
$ 1. Prove\ that\ \left|A\times C\right|=\left|B \times D\right|$
$ 2. \left|A^{n}\right|= \left|B^{n}\right| $
$ 3.\left|P\left(A\right)\right|=\left|P\left(B\righ t)\right| $
With P i mean the power set

thx guys

**JaneBennet** · January 24th 2008, 03:12 PM

$|A|=|B|\ \mbox{and}\ |C|=|D|\ \Rightarrow\ \exists\ \mbox{bijections}\ \mathrm{f}:A\to B\ \mbox{and}\ \mathrm{g}:C\to D$ . Now show that the following functions are bijections.

(1) $\mathrm{h}:A\times C\to B\times D,\ \mathrm{h}(a,c)=(\mathrm{f}(a),\mathrm{g}(c))$

(2) $\mathrm{h}:A^n\to B^n,\ \mathrm{h}(a_1,\dots,a_n)=(\mathrm{f}(a_1),\ldots\ ,\mathrm{f}(a_n))$

(2) $\mathrm{h}:\mathcal{P}(A)\to\mathcal{P}(B),\ \mathrm{h}(X)=\{\mathrm{f}(a):a\in X\}$

**Plato** · January 24th 2008, 03:15 PM

From the given, there are bijections: $f:A \leftrightarrow B\,\& \,g:C \leftrightarrow D$ .
Now define a function $\Phi :\left( {A \times C} \right) \leftrightarrow \left( {B \times D} \right),\,\,\Phi (a,c) = \left( {f(a),g(c)} \right)$ .
Prove that $\Phi$ is a bijection.

I have no idea what the notation in part b means.

For part c, use the $\chi$ function: $\chi _E (x) = \left\{ {\begin{array}{lr} 1 & {x \in E} \\ 0 & {x \notin {\rm E}} \\ \end{array}} \right.$

**Jhevon** · January 24th 2008, 03:17 PM

Originally Posted by kuntah

Hint: $|A \times C| = |A||C|$

2. $\left|A^{n}\right|= \left|B^{n}\right| $

Hint: $\left| A^n \right| = | \underbrace{A \times A \times A \cdots A}_{\mbox{n times}}| = \underbrace {|A||A||A| \cdots |A|}_{\mbox{n times}}$

3. $\left|P\left(A\right)\right|=\left|P\left(B\right) \right| $
With P i mean the power set

Hint: $|P(A)| = 2^{|A|}$

EDIT: ...ok, so both JaneBennet and Plato have different methods from mine, and theirs are similar, so chances are what I did was foolishness

. you may ignore this post. i'll leave it up for anyone to comment on it.

**ThePerfectHacker** · January 24th 2008, 04:00 PM

Originally Posted by Jhevon

EDIT: ...ok, so both JaneBennet and Plato have different methods from mine, and theirs are similar, so chances are what I did was foolishness

. you may ignore this post. i'll leave it up for anyone to comment on it.

The posters proved this for any sets in general. You assumed it were finite sets and applied finite numbers.

**Jhevon** · January 24th 2008, 04:04 PM

Originally Posted by ThePerfectHacker

The posters proved this for any sets in general. You assumed it were finite sets and applied finite numbers.

i thought so, thanks

Thread: Set equality

LinkBack

Thread Tools

Display

Set equality

Similar Math Help Forum Discussions

Predicate Calculus help (interpretation of equality in a model of equality)

Equality

Equality !!

Equality

equality

Search Tags