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The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each dense order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is dense in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either dense (and uniformly continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is dense unless $K$ is already complete.

Remark 1: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "dense" and "sequentially Cauchy" are precisely equivalent.

Remark 2: Cohn's completion is unusual because he only considers Cauchy sequences. However, it does coincide with the usual topological completion when the ordered field is metrizable. An ordered field is metrizable precisely if it has countable cofinality. When a field has uncountable cofinality, all Cauchy sequences are eventually constant and Cohn's completion trivializes.

The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each cofinal order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is cofinal in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either cofinal (and uniformly continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is cofinal unless $K$ is already complete.

Remark 1: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "cofinal" and "sequentially Cauchy" are precisely equivalent.

Remark 2: Cohn's completion is unusual because he only considers Cauchy sequences. However, it does coincide with the usual topological completion when the ordered field is metrizable. An ordered field is metrizable precisely if it has countable cofinality. When a field has uncountable cofinality, all Cauchy sequences are eventually constant and Cohn's completion trivializes.

The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each dense order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is dense in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either dense (and continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is dense unless $K$ is already complete.

Remark: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "dense" and "sequentially Cauchy" are precisely equivalent.

The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each dense order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is dense in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either dense (and uniformly continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is dense unless $K$ is already complete.

Remark 1: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "dense" and "sequentially Cauchy" are precisely equivalent.

Remark 2: Cohn's completion is unusual because he only considers Cauchy sequences. However, it does coincide with the usual topological completion when the ordered field is metrizable. An ordered field is metrizable precisely if it has countable cofinality. When a field has uncountable cofinality, all Cauchy sequences are eventually constant and Cohn's completion trivializes.

The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each dense order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is dense in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either dense (and continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is dense unless $K$ is already complete.

Remark: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy. This is the only case where one can remove the word dense from the theorem, in all other cases "sequentially Cauchy" is equivalent to dense.

The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each dense order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is dense in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either dense (and continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is dense unless $K$ is already complete.

Remark: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "dense" and "sequentially Cauchy" are precisely equivalent.