Return to Answer

Let's suppose the well-order of $A$ and $B$ have order-types $p$ (assume limit for simplicity) and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(i)=r \in A+B$, then we can write $first(r)=\min\{x\in A\,|\, \exists y \in B \, (x+y=r) \}$ and $second(r)=r-first(r)$. The specific definition of $first(t)$ is justified because $A$ is well-ordered. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.

Let's suppose the well-order of $A$ and $B$ have order-types $p$ (assume limit for simplicity) and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(i)=r \in A+B$, then we can write $first(r)=\min\{x\in A\,|\, \exists y \in B \, (x+y=r) \}$ and $second(r)=r-first(r)$. The specific definition of $first(r)$ is justified because $A$ is well-ordered. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.

Let's suppose the well-order of $A$ and $B$ have order-types $p$ (assume limit for simplicity) and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(x)=a_i+b_j$ then we can write $first(f(x))$ and $second(f(x))$ to denote $a_i$ and $b_j$ respectively. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.

Let's suppose the well-order of $A$ and $B$ have order-types $p$ (assume limit for simplicity) and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(i)=r \in A+B$, then we can write $first(r)=\min\{x\in A\,|\, \exists y \in B \, (x+y=r) \}$ and $second(r)=r-first(r)$. The specific definition of $first(t)$ is justified because $A$ is well-ordered. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.

Let's suppose the well-order of $A$ and $B$ have order-types $p$ and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(x)=a_i+b_j$ then we can write $first(f(x))$ and $second(f(x))$ to denote $a_i$ and $b_j$ respectively. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.

Let's suppose the well-order of $A$ and $B$ have order-types $p$ (assume limit for simplicity) and $q$ respectively. So we use the notation $a_i$ ($i<p$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<p$. Similarly, we use $b_i$ ($i<q$) to denote the $i$-th element of $B$.

Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<p$ and $j<q$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.

So if we have $f(x)=a_i+b_j$ then we can write $first(f(x))$ and $second(f(x))$ to denote $a_i$ and $b_j$ respectively. Now we can define $\alpha_0=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha)\}$. Let's denote $n_0 \in \mathbb{N}$ as the "last" value for which $first(f(n_0))=a_{\alpha_0}$.

Now we define $\alpha_1=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_0)\}$. Let's denote $n_1 \in \mathbb{N}$ as the "last" value for which $first(f(n_1))=a_{\alpha_1}$. Now because of $\alpha_1>\alpha_0$, we should get $a_{\alpha_1}>a_{\alpha_0}$ and hence $second(f(n_1))<second(f(n_0))$.

So it seems to me that when we define $\alpha_2=\min\{\alpha \in p \,|\, \exists x \in \mathbb{N}(first(f(x))=a_\alpha \, \wedge \,x>n_1)\}$ and $n_2$ as the "last" value for which $first(f(n_2))=a_{\alpha_2}$, then we should have similarly $second(f(n_2))<second(f(n_1))$. The last inequality is supposed to follow from $a_{\alpha_2}>a_{\alpha_1}$ (because $\alpha_2>\alpha_1$).

The previous paragraph seems to be suggestive of defining $\alpha_i$, $n_i$ generally for an arbitrary natural number $i$ and then creating an infinite descent for the second components:$......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))$. However, this descent goes against the assumption of $B$ being a well-ordered set.

Few clarification points:

(1) At some points comparison relation $<$ is used for real numbers and at some points it is used for ordinals.

(2) We can think of $first$ as a function $first:A+B \rightarrow A$. Similarly, we can think of $second$ as a function $second:A+B \rightarrow B$.

(3) If we have $first(f(i))=a_\alpha$ for some specific natural number $i$ and some specific ordinal $\alpha$, then there can't exist arbitrarily large natural numbers $j>i$ such that $first(f(j))=a_\alpha$.