If a sequence $(x_n)$ belongs to the closure of $A$ then there exist a sequence $(y_n)$ such that...












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I know that if, a point $x$ belongs to the closure of $A$ then exist a sequence in $A$ that converges to $x$.
But this is true for a sequence of points?










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  • 1




    $begingroup$
    $x_n=y_n$.{}{}{}{}{}{}
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 20:18
















0












$begingroup$


I know that if, a point $x$ belongs to the closure of $A$ then exist a sequence in $A$ that converges to $x$.
But this is true for a sequence of points?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $x_n=y_n$.{}{}{}{}{}{}
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 20:18














0












0








0





$begingroup$


I know that if, a point $x$ belongs to the closure of $A$ then exist a sequence in $A$ that converges to $x$.
But this is true for a sequence of points?










share|cite|improve this question











$endgroup$




I know that if, a point $x$ belongs to the closure of $A$ then exist a sequence in $A$ that converges to $x$.
But this is true for a sequence of points?







real-analysis






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edited Jan 12 at 20:53









rtybase

11.4k31533




11.4k31533










asked Jan 12 at 20:14









Gabriel CorrêaGabriel Corrêa

143




143








  • 1




    $begingroup$
    $x_n=y_n$.{}{}{}{}{}{}
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 20:18














  • 1




    $begingroup$
    $x_n=y_n$.{}{}{}{}{}{}
    $endgroup$
    – hamam_Abdallah
    Jan 12 at 20:18








1




1




$begingroup$
$x_n=y_n$.{}{}{}{}{}{}
$endgroup$
– hamam_Abdallah
Jan 12 at 20:18




$begingroup$
$x_n=y_n$.{}{}{}{}{}{}
$endgroup$
– hamam_Abdallah
Jan 12 at 20:18










1 Answer
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$begingroup$

If $x_{1} in overline{A}$, there is a sequence $(y_{n,1}) subset A$ such that $y_{n,1} to x_{1}$. If $x_{2} in overline{A}$, there is a sequence $(y_{n,2}) subset A$ such that $y_{n,2} to x_{2}$. The same idea works for every $x_{n}$.



Now, since $y_{n,1} to x_{1}$, take $epsilon = 1/1$ and so, there is $k_{1} in mathbb{N}$ such that ${n,1}> k_{1}$ implies $|y_{n,1}-x_{1}| < 1/1$. Also, since $y_{n,2} to x_{2}$, take $epsilon = 1/2$ and so, there is $k_{2} in mathbb{N}$ such that $|y_{n,2}-x_{2}| < 1/2$ for some $k_{2} in mathbb{N}$ such that ${n,2} > k_{2}$. This works for every $x_{n}$.



Can you complete?






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  • $begingroup$
    Obrigado Lucas! Mas, devo fazer indução ?
    $endgroup$
    – Gabriel Corrêa
    Jan 12 at 21:16











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$begingroup$

If $x_{1} in overline{A}$, there is a sequence $(y_{n,1}) subset A$ such that $y_{n,1} to x_{1}$. If $x_{2} in overline{A}$, there is a sequence $(y_{n,2}) subset A$ such that $y_{n,2} to x_{2}$. The same idea works for every $x_{n}$.



Now, since $y_{n,1} to x_{1}$, take $epsilon = 1/1$ and so, there is $k_{1} in mathbb{N}$ such that ${n,1}> k_{1}$ implies $|y_{n,1}-x_{1}| < 1/1$. Also, since $y_{n,2} to x_{2}$, take $epsilon = 1/2$ and so, there is $k_{2} in mathbb{N}$ such that $|y_{n,2}-x_{2}| < 1/2$ for some $k_{2} in mathbb{N}$ such that ${n,2} > k_{2}$. This works for every $x_{n}$.



Can you complete?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Obrigado Lucas! Mas, devo fazer indução ?
    $endgroup$
    – Gabriel Corrêa
    Jan 12 at 21:16
















0












$begingroup$

If $x_{1} in overline{A}$, there is a sequence $(y_{n,1}) subset A$ such that $y_{n,1} to x_{1}$. If $x_{2} in overline{A}$, there is a sequence $(y_{n,2}) subset A$ such that $y_{n,2} to x_{2}$. The same idea works for every $x_{n}$.



Now, since $y_{n,1} to x_{1}$, take $epsilon = 1/1$ and so, there is $k_{1} in mathbb{N}$ such that ${n,1}> k_{1}$ implies $|y_{n,1}-x_{1}| < 1/1$. Also, since $y_{n,2} to x_{2}$, take $epsilon = 1/2$ and so, there is $k_{2} in mathbb{N}$ such that $|y_{n,2}-x_{2}| < 1/2$ for some $k_{2} in mathbb{N}$ such that ${n,2} > k_{2}$. This works for every $x_{n}$.



Can you complete?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Obrigado Lucas! Mas, devo fazer indução ?
    $endgroup$
    – Gabriel Corrêa
    Jan 12 at 21:16














0












0








0





$begingroup$

If $x_{1} in overline{A}$, there is a sequence $(y_{n,1}) subset A$ such that $y_{n,1} to x_{1}$. If $x_{2} in overline{A}$, there is a sequence $(y_{n,2}) subset A$ such that $y_{n,2} to x_{2}$. The same idea works for every $x_{n}$.



Now, since $y_{n,1} to x_{1}$, take $epsilon = 1/1$ and so, there is $k_{1} in mathbb{N}$ such that ${n,1}> k_{1}$ implies $|y_{n,1}-x_{1}| < 1/1$. Also, since $y_{n,2} to x_{2}$, take $epsilon = 1/2$ and so, there is $k_{2} in mathbb{N}$ such that $|y_{n,2}-x_{2}| < 1/2$ for some $k_{2} in mathbb{N}$ such that ${n,2} > k_{2}$. This works for every $x_{n}$.



Can you complete?






share|cite|improve this answer









$endgroup$



If $x_{1} in overline{A}$, there is a sequence $(y_{n,1}) subset A$ such that $y_{n,1} to x_{1}$. If $x_{2} in overline{A}$, there is a sequence $(y_{n,2}) subset A$ such that $y_{n,2} to x_{2}$. The same idea works for every $x_{n}$.



Now, since $y_{n,1} to x_{1}$, take $epsilon = 1/1$ and so, there is $k_{1} in mathbb{N}$ such that ${n,1}> k_{1}$ implies $|y_{n,1}-x_{1}| < 1/1$. Also, since $y_{n,2} to x_{2}$, take $epsilon = 1/2$ and so, there is $k_{2} in mathbb{N}$ such that $|y_{n,2}-x_{2}| < 1/2$ for some $k_{2} in mathbb{N}$ such that ${n,2} > k_{2}$. This works for every $x_{n}$.



Can you complete?







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share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 20:44









Lucas CorrêaLucas Corrêa

1,5751421




1,5751421












  • $begingroup$
    Obrigado Lucas! Mas, devo fazer indução ?
    $endgroup$
    – Gabriel Corrêa
    Jan 12 at 21:16


















  • $begingroup$
    Obrigado Lucas! Mas, devo fazer indução ?
    $endgroup$
    – Gabriel Corrêa
    Jan 12 at 21:16
















$begingroup$
Obrigado Lucas! Mas, devo fazer indução ?
$endgroup$
– Gabriel Corrêa
Jan 12 at 21:16




$begingroup$
Obrigado Lucas! Mas, devo fazer indução ?
$endgroup$
– Gabriel Corrêa
Jan 12 at 21:16


















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