Completion of surreal subfields












7














Let $kappa$ be a regular uncountable ordinal. Let $No(kappa)$ denote the field of surreal numbers of birthdate $ < kappa$.



In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(kappa)$ is isomorphic to $mathbb{R}((x))^{No(kappa)}_{<kappa}$ which is the subfield of Hahn series of length $< kappa$ over $mathbb{R}$ with value group $No(kappa)$.



Let $S$ be the subset of $mathbb{R}((x))^{No(kappa)} = mathbb{R}((x))^{No(kappa)}_{<{kappa}^+}$ of Hahn series of either length $<kappa$ or of length $kappa$ whose $kappa$-sequence of exponents is cofinal in $No(kappa)$.



It is not too difficult to see that $S$ is stable under $+$ and $-$.



I wonder if it is a subfield of $mathbb{R}((x))^{No(kappa)}$, in which case it would be an example of completion of $No(kappa)$. Does anyone know how to prove/disprove this?










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  • 2




    @meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
    – nombre
    Nov 29 '18 at 9:05












  • Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
    – meowzz
    Dec 3 '18 at 3:10
















7














Let $kappa$ be a regular uncountable ordinal. Let $No(kappa)$ denote the field of surreal numbers of birthdate $ < kappa$.



In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(kappa)$ is isomorphic to $mathbb{R}((x))^{No(kappa)}_{<kappa}$ which is the subfield of Hahn series of length $< kappa$ over $mathbb{R}$ with value group $No(kappa)$.



Let $S$ be the subset of $mathbb{R}((x))^{No(kappa)} = mathbb{R}((x))^{No(kappa)}_{<{kappa}^+}$ of Hahn series of either length $<kappa$ or of length $kappa$ whose $kappa$-sequence of exponents is cofinal in $No(kappa)$.



It is not too difficult to see that $S$ is stable under $+$ and $-$.



I wonder if it is a subfield of $mathbb{R}((x))^{No(kappa)}$, in which case it would be an example of completion of $No(kappa)$. Does anyone know how to prove/disprove this?










share|cite|improve this question


















  • 2




    @meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
    – nombre
    Nov 29 '18 at 9:05












  • Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
    – meowzz
    Dec 3 '18 at 3:10














7












7








7


2





Let $kappa$ be a regular uncountable ordinal. Let $No(kappa)$ denote the field of surreal numbers of birthdate $ < kappa$.



In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(kappa)$ is isomorphic to $mathbb{R}((x))^{No(kappa)}_{<kappa}$ which is the subfield of Hahn series of length $< kappa$ over $mathbb{R}$ with value group $No(kappa)$.



Let $S$ be the subset of $mathbb{R}((x))^{No(kappa)} = mathbb{R}((x))^{No(kappa)}_{<{kappa}^+}$ of Hahn series of either length $<kappa$ or of length $kappa$ whose $kappa$-sequence of exponents is cofinal in $No(kappa)$.



It is not too difficult to see that $S$ is stable under $+$ and $-$.



I wonder if it is a subfield of $mathbb{R}((x))^{No(kappa)}$, in which case it would be an example of completion of $No(kappa)$. Does anyone know how to prove/disprove this?










share|cite|improve this question













Let $kappa$ be a regular uncountable ordinal. Let $No(kappa)$ denote the field of surreal numbers of birthdate $ < kappa$.



In Fields of surreal numbers and exponentiation (2000), P. Ehrlich and L.v.d. Dries proved that $No(kappa)$ is isomorphic to $mathbb{R}((x))^{No(kappa)}_{<kappa}$ which is the subfield of Hahn series of length $< kappa$ over $mathbb{R}$ with value group $No(kappa)$.



Let $S$ be the subset of $mathbb{R}((x))^{No(kappa)} = mathbb{R}((x))^{No(kappa)}_{<{kappa}^+}$ of Hahn series of either length $<kappa$ or of length $kappa$ whose $kappa$-sequence of exponents is cofinal in $No(kappa)$.



It is not too difficult to see that $S$ is stable under $+$ and $-$.



I wonder if it is a subfield of $mathbb{R}((x))^{No(kappa)}$, in which case it would be an example of completion of $No(kappa)$. Does anyone know how to prove/disprove this?







field-theory ordinals ordered-fields surreal-numbers






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 '16 at 2:33









nombre

2,634913




2,634913








  • 2




    @meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
    – nombre
    Nov 29 '18 at 9:05












  • Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
    – meowzz
    Dec 3 '18 at 3:10














  • 2




    @meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
    – nombre
    Nov 29 '18 at 9:05












  • Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
    – meowzz
    Dec 3 '18 at 3:10








2




2




@meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
– nombre
Nov 29 '18 at 9:05






@meowzz: There as been an answer by Philip Ehrlich to a similar question of mine [here][1]. It is actually standard in valuation theory that the Cauchy-completion of a (non archimedean) valued field can naturally be constructed in this way. [1]: mathoverflow.net/questions/237769/completing-class-sized-fields/…
– nombre
Nov 29 '18 at 9:05














Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
– meowzz
Dec 3 '18 at 3:10




Noted. If nothing else, you could post (your version of?) the answer here & I'd be happy to award the bounty to you as you have a lot of other answers that I have found incredibly helpful.
– meowzz
Dec 3 '18 at 3:10










1 Answer
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+50









So as said in the comments, this follows from the following general result in valuation theory:



Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.



For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.



Corollary 3.2.18 in ADH:



If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.



In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.





Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.



Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.



Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.



Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.






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    So as said in the comments, this follows from the following general result in valuation theory:



    Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.



    For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.



    Corollary 3.2.18 in ADH:



    If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.



    In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.





    Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.



    Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.



    Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.



    Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.






    share|cite|improve this answer




























      1





      +50









      So as said in the comments, this follows from the following general result in valuation theory:



      Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.



      For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.



      Corollary 3.2.18 in ADH:



      If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.



      In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.





      Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.



      Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.



      Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.



      Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.






      share|cite|improve this answer


























        1





        +50







        1





        +50



        1




        +50




        So as said in the comments, this follows from the following general result in valuation theory:



        Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.



        For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.



        Corollary 3.2.18 in ADH:



        If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.



        In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.





        Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.



        Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.



        Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.



        Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.






        share|cite|improve this answer














        So as said in the comments, this follows from the following general result in valuation theory:



        Let $F$ be a field, let $(G,+)$ be a non trivial ordered group, and let $F[[varepsilon^G]]$ be the field of Hahn series with value group $G$ and coefficients in $F$. This is equipped with the uniform structure and topology induced by the standard valuation $v$ whose valuation ring is the set of series whose support is a subset of $G^{geq 0}$.



        For $f in F[[varepsilon^G]]$ and $g in G$, we let $f|_g$ denote the series with support $operatorname{supp} f cap G^{> g}$ which coincides with $f$ on this set. This is a truncation of $f$ as a series.



        Corollary 3.2.18 in ADH:



        If $T$ is a subfield of $F[[varepsilon^G]]$ which is stable under truncation and contains $varepsilon^G$, then the set $overline{T}:={f in F[[varepsilon^G]]:forall g in G, f|_g in T}$ is a completion of $T$, that is, a maximal dense valued field extension of $T$.



        In the case when $F$ is ordered, the same definition gives a maximal dense ordered field extension of $T$, that is, the Cauchy-completion of $T$ as an ordered field.





        Let's apply this to the case $F=mathbb{R}$, $G=mathbf{No}(kappa)$ and $T=mathbf{No}(kappa)$. We have $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$ which is stable under truncation in $mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa^+}$ and contains $varepsilon^{mathbf{No}(kappa)}$.



        Let $f in overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}$. If $operatorname{supp} f$ is cofinal in $mathbf{No}(kappa)$, then let $alpha$ denote its order type and assume $kappa< alpha$. The ordinal $alpha$ must be limit so $kappa+1<alpha$. By the definition of the completion, we have $f|_{x_{kappa+1}} in mathbf{No}(kappa)$ where $x_{kappa+1}$ is the $(kappa+1)$-th element of $operatorname{supp} f$. But the order type of $operatorname{supp} f|_{x_{kappa+1}}$ is $kappa$, which contradicts the identity $mathbf{No}(kappa)=mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}$. So $alphaleqkappa$, and thus $f in S$. Else $operatorname{supp} f$ is has an upper bound $x$ in $mathbf{No}(kappa)$ and then $f|_{x+1}=f in mathbf{No}(kappa)$ so $f in S$. This proves the inclusion $overline{mathbf{No}(kappa)} subseteq S$.



        Conversely we have $mathbf{No}(kappa) subset overline{mathbf{No}(kappa)}$. For $f in mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{leq kappa}$ whose support is cofinal in $mathbf{No}$, the order type $alpha$ of $operatorname{supp} f$ satisfies $kappa=operatorname{cof}(kappa)leqalphaleq kappa$ so $alpha=kappa$. For $x in mathbf{No}(kappa)$, there is $y in operatorname{supp} f$ with $x<y$, so $operatorname{supp} f|_x<operatorname{supp} f|_yleq kappa$. Thus $f|_x$ lies in $mathbf{No}(kappa)$. We deduce that $f in overline{mathbf{No}(kappa)}$.



        Hence $S=overline{mathbb{R}[[varepsilon^{mathbf{No}(kappa)}]]_{<kappa}}=overline{mathbf{No}(kappa)}$ as desired.







        share|cite|improve this answer














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        edited Dec 3 '18 at 21:19

























        answered Dec 3 '18 at 19:21









        nombre

        2,634913




        2,634913






























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