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-I claim that $N_1$ doesn't have a real closure. Let's assume it does. Then, by theorem 19 in $[1]$, this real closure, and therefore $N_1$ itself, embeds in $No$. Let $\sigma$ be such an embedding. $\sigma(N_1)$ must have an upper bound in $No$ because else $\sigma(X)$ would be a surreal whose sequence of powers is cofinal in $No$, which is impossible for various reasons. So $\sigma(N_1)$ is a subset of the convex class $C$ generated by $No(\kappa)$ for some uncountable cardinal $\kappa$ which is a strict upper bound of $\sigma(N_1)$. We can cut $C$ in less than $2^{\kappa}$ many disjoint intervals of diameter $\frac{1}{\kappa}$ using the classical integer part of $No(\kappa)$ (Conway ingeters): write $C = \bigsqcup \limits_{a \in Oz \cap No(\kappa)} [\frac{1}{\kappa}a;\frac{1}{\kappa}(a+1)[_{No}$. Now, taking a subset of $N_1$ of size $\geq {2^{\kappa}}^+$ and using the pigeonhole principle with $\sigma$, we see that one of those intervals must contain two elements of $\sigma(N_1)$, but then the inverse of their absolute difference lies outside $C$ because it is greater than $\kappa$: a constradiction.

-I claim that $N_1$ doesn't have a real closure. Let's assume it does. Then, by theorem 19 in $[1]$, this real closure, and therefore $N_1$ itself, embeds in $No$. Let $\sigma$ be such an embedding. $\sigma(N_1)$ must have an upper bound in $No$ because else $\sigma(X)$ would be a surreal whose sequence of powers is cofinal in $No$, which is impossible for various reasons. So $\sigma(N_1)$ is a subset of the convex class $C$ generated by $No({\kappa}^+)$ for some cardinal $\kappa$ which is a strict upper bound of $\sigma(N_1)$. We can cut $C$ in less than $2^{{\kappa}^+}$ many disjoint intervals of diameter $\frac{1}{\kappa}$ using the classical integer part of $No({\kappa}^+)$ (Conway ingeters): write $C = \bigsqcup \limits_{a \in Oz \cap No({\kappa}^+)} [\frac{1}{\kappa}a;\frac{1}{\kappa}(a+1)[_{No}$. Now, taking a subset of $N_1$ of size $\geq (2^{{\kappa}^+})^+$ and using the pigeonhole principle with $\sigma$, we see that one of those intervals must contain two elements of $\sigma(N_1)$, but then the inverse of their absolute difference isn't in $\sigma(N_1)$ because it is greater than $\kappa$: a constradiction.

-I claim that $N_1$ doesn't have a real closure. Let's assume it does. Then, by theorem 19 in $[1]$, this real closure, and therefore $N_1$ itself, embed in $No$. Let $\sigma$ be such an embedding. $\sigma(N_1)$ must have an upper bound in $No$ because else $\sigma(X)$ would be a surreal whose sequence of powers is cofinal in $No$, which is impossible for various reasons. So $\sigma(N_1)$ is a subset of the convex class $C$ generated by $No(\kappa)$ for some uncountable cardinal $\kappa$ which is a strict upper bound of $\sigma(N_1)$. We can cut $C$ in less than $2^{\kappa}$ many disjoint interval of diameter $\frac{1}{\kappa}$ using the classical integer part of $No(\kappa)$ (Conway ingeters). Now, because taking a subset of $N_1$ of size $\geq {2^{\kappa}}^+$ and using the pigeonhole principle with $\sigma$, we see that one of those intervals must contain two elements of $\sigma(N_1)$, but then the inverse of their difference lies outside $C$ because it is greater than $\kappa$: a constradiction.

-I claim that $N_1$ doesn't have a real closure. Let's assume it does. Then, by theorem 19 in $[1]$, this real closure, and therefore $N_1$ itself, embeds in $No$. Let $\sigma$ be such an embedding. $\sigma(N_1)$ must have an upper bound in $No$ because else $\sigma(X)$ would be a surreal whose sequence of powers is cofinal in $No$, which is impossible for various reasons. So $\sigma(N_1)$ is a subset of the convex class $C$ generated by $No(\kappa)$ for some uncountable cardinal $\kappa$ which is a strict upper bound of $\sigma(N_1)$. We can cut $C$ in less than $2^{\kappa}$ many disjoint intervals of diameter $\frac{1}{\kappa}$ using the classical integer part of $No(\kappa)$ (Conway ingeters): write $C = \bigsqcup \limits_{a \in Oz \cap No(\kappa)} [\frac{1}{\kappa}a;\frac{1}{\kappa}(a+1)[_{No}$. Now, taking a subset of $N_1$ of size $\geq {2^{\kappa}}^+$ and using the pigeonhole principle with $\sigma$, we see that one of those intervals must contain two elements of $\sigma(N_1)$, but then the inverse of their absolute difference lies outside $C$ because it is greater than $\kappa$: a constradiction.

-Every ordered field $F$ has a real closure, that is a unique real closed (and algebraic) extension $(\mathcal{R}(F),\rho)$ such that whenever $F$ embeds in an algebraic extension $(F',\lambda)$, there is an embedding $\varphi: \mathcal{R}(F) \rightarrow F'$ such that $\varphi \circ \rho = \lambda$.

-Every ordered field $F$ has a completion, that is a unique complete (and dense) extension $(\widetilde{F},\mu)$ such that whenever $F$ embeds in a cofinal extension $(F',\lambda)$, there is an embedding $\varphi: \widetilde{F} \rightarrow F'$ such that $\varphi \circ \mu = \lambda$.

-Every ordered field $F$ has a real closure, that is a unique real closed (and algebraic) extension $(\mathcal{R}(F),\rho)$ such that whenever $F$ embeds in a real closed extension $(F',\lambda)$, there is an embedding $\varphi: \mathcal{R}(F) \rightarrow F'$ such that $\varphi \circ \rho = \lambda$.

-Every ordered field $F$ has a completion, that is a unique complete (and dense) extension $(\widetilde{F},\mu)$ such that whenever $F$ embeds in a cofinal complete extension $(F',\lambda)$, there is an embedding $\varphi: \widetilde{F} \rightarrow F'$ such that $\varphi \circ \mu = \lambda$.