Show that if $(x_1,x_2)$ is defined to be $\{\{x_1\},\{x_1,x_2\}\}$ then $(x_1,x_2)=(y_1,y_2)$ iff $x_1=y_1$ and $x_2=y_2$

Question

My Work:

If you take the cartesian product of any set with two arbitrary elements $a$ and $b$, and the resulting set is $\{\{x_1\},\{x_1,x_2\}\}$, then the only possible values for $a$ and $b$ are $x_1$ and $x_2$ by definition of the cartesian product.

This answer seems overly trivial and I think I'm doing something wrong. Please help!

I think your definition is wrong. Shouldn't it be ${{x_1}, {x_1, x_2}}$? — Clarinetist, Feb 09 '16 at 03:18
The definition of cartesian product $A\times B$ is "the set of pairs $(a,b)$ with $a\in A$ and $b\in B$", therefore your argument is most likely circular. — , Feb 09 '16 at 07:28

fred · Answer 1 · 2016-02-09T18:28:17.740

Lemma: Denote the cardinality of a set,$A$, by $|A|$. Suppose $A$ and $B$ are sets with $|A| = 2$ and $|B|=1$, then $A\not = B$.

Proof: Let $A=\{a_1,a_2\}$ with $a_1 \not = a_2$, and $B=\{b_1\}$, then if $b_1\not \in A$ we have $A\not = B$ and we are done, so assume that $b_1\in A$. Without loss of generality we can say $b_1=a_1$. But in this case we have $a_2\not\in B$ so $A\not = B$.

First of all, it is easy to see that if $x_1 = y_1$ and $x_2 = y_2$ we have $$ \left\{ \{x_1\}, \{x_1,x_2\}\right\} = \left\{ \{y_1\}, \{y_1,y_2\}\right\}, $$ so we have the backward implication. For the forward implication we have two cases, for the case $x_1=x_2$ we have

\begin{align*} (x_1,x_2) = (y_1,y_2)& \Leftrightarrow (x_1,x_1) = (y_1,y_2)\\ &\Leftrightarrow \left\{\{x_1\}, \{x_1,x_1\} \right\} = \left\{ \{ y_1\}, \{y_1,y_2\} \right\}\\ &\Leftrightarrow \left\{ \{x_1\}, \{x_1\} \right\} = \left\{\{y_1\}, \{y_1,y_2\} \right\}\\ &\Leftrightarrow \left\{ \{x_1\} \right\} = \left\{ \{y_1\}, \{y_1,y_2\} \right\}\\ &\Rightarrow x_1 = y_1\ \mathrm{and\ } \{x_1\} = \{y_1,y_2\}\\ &\Rightarrow x_1 = y_1\ \mathrm{and\ } x_1=y_1=y_2\\ &\Rightarrow x_1 = y_1\ \mathrm{and\ } x_2=y_2. \end{align*}

For the case $x_1\not= x_2$ we have \begin{align*} (x_1,x_2) = (y_1,y_2) &\Leftrightarrow \left\{ \{ x_1\}, \{x_1,x_2\} \right\}= \left\{ \{y_1\}, \{y_1,y_2\} \right\}\\ &\Leftrightarrow \{x_1\} = \{y_1\}\ \mathrm{and\ } \{x_1,x_2\} = \{y_1,y_2\}\\ &\Leftrightarrow x_1 = y_1\ \mathrm{and\ } \{x_1,x_2\} = \{y_1,y_2\}\\ &\Rightarrow x_1 = y_1\ \mathrm{and\ } \{x_1,x_2\} = \{x_1,y_2\}\\ &\Rightarrow x_1 = y_1\ \mathrm{and\ } x_2 = y_2. \end{align*}

I think you should explicitly deal with the case $${x_1}={y_1,y_2}\text{ and } {x_1,x_2}={y_1}$$ (because it's the only technical issue of the proof) — , Feb 09 '16 at 08:04
@G.Sassatelli I think I have addressed your concerns, I still haven't done every step in the most explicit detail. We are able to discard the case you mentioned if we assume $x_1\not = x_2$. I address the case $x_1 = x_2$ separately. — fred, Feb 09 '16 at 14:20
@fred Not really. I know that the case I mentioned is not possible when $x_1\ne x_2$ but, again: proving it is crucial, because at this point of the theory you are yet to define the meaning of "having exactly $2$ elements". — , Feb 09 '16 at 15:39

Show that if $(x_1,x_2)$ is defined to be $\{\{x_1\},\{x_1,x_2\}\}$ then $(x_1,x_2)=(y_1,y_2)$ iff $x_1=y_1$ and $x_2=y_2$

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