Well Orderings and Transfinite Induction - Set Theory

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Question

Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

Answer

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$f:A_1\rightarrow A_2$ and $g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it .

$t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, can be defined as the function,

$\t:A_1\times B_1\rightarrow A_2\times B_2 \t(a,b)=(f(a),g(b))$

To show is a well defined, one-to-one, function on $A_1\times B_1$ since and are one-to-one, the following is performed.

$t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

$\\Rightarrow f(a)=f(u), g(b)=g(v) \\Rightarrow a=u, b=v \\Rightarrow (a,b)=(u,v)$

Thus, is one-to-one.

Now to prove in onto $A_2\times B_2$ if $(a,b)\ \epsilon\ A_2\times B_2$

for some

$u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

proving is onto $A_2 \times B_2$ .

Lastly prove ordering.

$(a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left{\begin{matrix} a_1<_aa_2\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} f(a_1)<_{A}f(a_2)\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} (f(a_1),g(b_1))<_{A\times B}(f(a_2),g(b_2))\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving is isomorphic. Therefore the statement is true.

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Question

Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

Answer

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$f:A_1\rightarrow A_2$ and $g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it .

$t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, can be defined as the function,

$\t:A_1\times B_1\rightarrow A_2\times B_2 \t(a,b)=(f(a),g(b))$

To show is a well defined, one-to-one, function on $A_1\times B_1$ since and are one-to-one, the following is performed.

$t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

$\\Rightarrow f(a)=f(u), g(b)=g(v) \\Rightarrow a=u, b=v \\Rightarrow (a,b)=(u,v)$

Thus, is one-to-one.

Now to prove in onto $A_2\times B_2$ if $(a,b)\ \epsilon\ A_2\times B_2$

for some

$u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

proving is onto $A_2 \times B_2$ .

Lastly prove ordering.

$(a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left{\begin{matrix} a_1<_aa_2\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} f(a_1)<_{A}f(a_2)\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} (f(a_1),g(b_1))<_{A\times B}(f(a_2),g(b_2))\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving is isomorphic. Therefore the statement is true.

Compare your answer with the correct one above

Question

Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

Answer

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$f:A_1\rightarrow A_2$ and $g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it .

$t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, can be defined as the function,

$\t:A_1\times B_1\rightarrow A_2\times B_2 \t(a,b)=(f(a),g(b))$

To show is a well defined, one-to-one, function on $A_1\times B_1$ since and are one-to-one, the following is performed.

$t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

$\\Rightarrow f(a)=f(u), g(b)=g(v) \\Rightarrow a=u, b=v \\Rightarrow (a,b)=(u,v)$

Thus, is one-to-one.

Now to prove in onto $A_2\times B_2$ if $(a,b)\ \epsilon\ A_2\times B_2$

for some

$u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

proving is onto $A_2 \times B_2$ .

Lastly prove ordering.

$(a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left{\begin{matrix} a_1<_aa_2\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} f(a_1)<_{A}f(a_2)\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} (f(a_1),g(b_1))<_{A\times B}(f(a_2),g(b_2))\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving is isomorphic. Therefore the statement is true.

Compare your answer with the correct one above

Question

Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

Answer

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$f:A_1\rightarrow A_2$ and $g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it .

$t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, can be defined as the function,

$\t:A_1\times B_1\rightarrow A_2\times B_2 \t(a,b)=(f(a),g(b))$

To show is a well defined, one-to-one, function on $A_1\times B_1$ since and are one-to-one, the following is performed.

$t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

$\\Rightarrow f(a)=f(u), g(b)=g(v) \\Rightarrow a=u, b=v \\Rightarrow (a,b)=(u,v)$

Thus, is one-to-one.

Now to prove in onto $A_2\times B_2$ if $(a,b)\ \epsilon\ A_2\times B_2$

for some

$u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

proving is onto $A_2 \times B_2$ .

Lastly prove ordering.

$(a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left{\begin{matrix} a_1<_aa_2\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} f(a_1)<_{A}f(a_2)\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} (f(a_1),g(b_1))<_{A\times B}(f(a_2),g(b_2))\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving is isomorphic. Therefore the statement is true.

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