Proving that if the dual space $X^*$ of a Banach space $X$ is separable , then $X$ is separable












2












$begingroup$


I have reading an answer post before , and there's something I do not understand .

For each $f_n$ in the unit ball of $X^*$ , why there exist $x_n in X$ with $|x_n | le1$ such that $f_n(x_n) ge frac{1}{2}$ .

I can show that for each $f_n$ , there exist $x in X$ , $f_n(x)=a neq 0$ . So let $x_n = frac{x}{a}$ , we get the inequality . But how to get the desired $x_n$ with $x_n$ also in the unit ball of $X$ ?






functional-analysis banach-spaces dual-spaces separable-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
    $endgroup$
    – Martin Sleziak
    Dec 8 '18 at 13:43
















2












$begingroup$


I have reading an answer post before , and there's something I do not understand .

For each $f_n$ in the unit ball of $X^*$ , why there exist $x_n in X$ with $|x_n | le1$ such that $f_n(x_n) ge frac{1}{2}$ .

I can show that for each $f_n$ , there exist $x in X$ , $f_n(x)=a neq 0$ . So let $x_n = frac{x}{a}$ , we get the inequality . But how to get the desired $x_n$ with $x_n$ also in the unit ball of $X$ ?






functional-analysis banach-spaces dual-spaces separable-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
    $endgroup$
    – Martin Sleziak
    Dec 8 '18 at 13:43














2












2








2





$begingroup$


I have reading an answer post before , and there's something I do not understand .

For each $f_n$ in the unit ball of $X^*$ , why there exist $x_n in X$ with $|x_n | le1$ such that $f_n(x_n) ge frac{1}{2}$ .

I can show that for each $f_n$ , there exist $x in X$ , $f_n(x)=a neq 0$ . So let $x_n = frac{x}{a}$ , we get the inequality . But how to get the desired $x_n$ with $x_n$ also in the unit ball of $X$ ?






functional-analysis banach-spaces dual-spaces separable-spaces







share|cite|improve this question











$endgroup$




I have reading an answer post before , and there's something I do not understand .

For each $f_n$ in the unit ball of $X^*$ , why there exist $x_n in X$ with $|x_n | le1$ such that $f_n(x_n) ge frac{1}{2}$ .

I can show that for each $f_n$ , there exist $x in X$ , $f_n(x)=a neq 0$ . So let $x_n = frac{x}{a}$ , we get the inequality . But how to get the desired $x_n$ with $x_n$ also in the unit ball of $X$ ?






functional-analysis banach-spaces dual-spaces separable-spaces




functional-analysis banach-spaces dual-spaces separable-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 13:39







J.Guo

















asked Dec 8 '18 at 13:33









J.GuoJ.Guo

3209




3209








  • 1




    $begingroup$
    I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
    $endgroup$
    – Martin Sleziak
    Dec 8 '18 at 13:43














  • 1




    $begingroup$
    I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
    $endgroup$
    – Martin Sleziak
    Dec 8 '18 at 13:43








1




1




$begingroup$
I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 13:43




$begingroup$
I think that both comments and the answer there explain that the proof as written there assumes that $f_n$ belong to the unit sphere, i.e., $|f_n|=1$. Using that, you can get $x_n$ such that $f_n(x_n)gefrac12$.
$endgroup$
– Martin Sleziak
Dec 8 '18 at 13:43










1 Answer
1






active

oldest

votes


















2












$begingroup$

That is false. It should read, the unit sphere, not the unit ball.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
    $endgroup$
    – J.Guo
    Dec 8 '18 at 13:48










  • $begingroup$
    Quite right : )
    $endgroup$
    – Ben W
    Dec 8 '18 at 13:49











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

That is false. It should read, the unit sphere, not the unit ball.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
    $endgroup$
    – J.Guo
    Dec 8 '18 at 13:48










  • $begingroup$
    Quite right : )
    $endgroup$
    – Ben W
    Dec 8 '18 at 13:49
















2












$begingroup$

That is false. It should read, the unit sphere, not the unit ball.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
    $endgroup$
    – J.Guo
    Dec 8 '18 at 13:48










  • $begingroup$
    Quite right : )
    $endgroup$
    – Ben W
    Dec 8 '18 at 13:49














2












2








2





$begingroup$

That is false. It should read, the unit sphere, not the unit ball.






share|cite|improve this answer









$endgroup$



That is false. It should read, the unit sphere, not the unit ball.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 13:42









Ben WBen W

2,274615




2,274615












  • $begingroup$
    Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
    $endgroup$
    – J.Guo
    Dec 8 '18 at 13:48










  • $begingroup$
    Quite right : )
    $endgroup$
    – Ben W
    Dec 8 '18 at 13:49


















  • $begingroup$
    Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
    $endgroup$
    – J.Guo
    Dec 8 '18 at 13:48










  • $begingroup$
    Quite right : )
    $endgroup$
    – Ben W
    Dec 8 '18 at 13:49
















$begingroup$
Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
$endgroup$
– J.Guo
Dec 8 '18 at 13:48




$begingroup$
Is it right that : $f $ belongs to the unit sphere means $|f | =1 $ and $f $ belongs to the unit ball means $| f | le 1$ ?
$endgroup$
– J.Guo
Dec 8 '18 at 13:48












$begingroup$
Quite right : )
$endgroup$
– Ben W
Dec 8 '18 at 13:49




$begingroup$
Quite right : )
$endgroup$
– Ben W
Dec 8 '18 at 13:49


















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