Show that $x_0 in overline{langle M rangle } $ if and only if $f(x_0) = 0, ; forall f in X^* : f|_M = 0$.












1












$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22
















1












$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22














1












1








1





$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$




Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem




real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 16:16









RebellosRebellos

15.4k31250




15.4k31250












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22


















  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22
















$begingroup$
The definition you state about linear hull is wrong by the way.
$endgroup$
– Shashi
Dec 17 '18 at 17:22




$begingroup$
The definition you state about linear hull is wrong by the way.
$endgroup$
– Shashi
Dec 17 '18 at 17:22










1 Answer
1






active

oldest

votes


















2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19











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1 Answer
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1 Answer
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active

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2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19
















2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19














2












2








2





$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$



$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 22:06

























answered Dec 17 '18 at 16:49









ShashiShashi

7,2951628




7,2951628












  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19


















  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19
















$begingroup$
Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
$endgroup$
– Rebellos
Dec 17 '18 at 17:59




$begingroup$
Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
$endgroup$
– Rebellos
Dec 17 '18 at 17:59












$begingroup$
@Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
$endgroup$
– Shashi
Dec 17 '18 at 18:00




$begingroup$
@Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
$endgroup$
– Shashi
Dec 17 '18 at 18:00












$begingroup$
I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
$endgroup$
– Rebellos
Dec 17 '18 at 18:04




$begingroup$
I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
$endgroup$
– Rebellos
Dec 17 '18 at 18:04












$begingroup$
@Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
$endgroup$
– Shashi
Dec 17 '18 at 18:06




$begingroup$
@Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
$endgroup$
– Shashi
Dec 17 '18 at 18:06




1




1




$begingroup$
@Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
$endgroup$
– Shashi
Dec 17 '18 at 18:19




$begingroup$
@Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
$endgroup$
– Shashi
Dec 17 '18 at 18:19


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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