If a sequence $(x_n)$ belongs to the closure of $A$ then there exist a sequence $(y_n)$ such that...
$begingroup$
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?
real-analysis
edited Jan 12 at 20:53
rtybase
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
$endgroup$
add a comment |
$begingroup$
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?
real-analysis
edited Jan 12 at 20:53
rtybase
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
$endgroup$
1
$begingroup$
$x_n=y_n$.{}{}{}{}{}{}
$endgroup$
– hamam_Abdallah
Jan 12 at 20:18
add a comment |
$begingroup$
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?
real-analysis
edited Jan 12 at 20:53
rtybase
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
$endgroup$
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
real-analysis
edited Jan 12 at 20:53
rtybase
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
edited Jan 12 at 20:53
rtybase
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
edited Jan 12 at 20:53
rtybase
11.4k31533
edited Jan 12 at 20:53
rtybase
11.4k31533
edited Jan 12 at 20:53
rtybase
11.4k31533
11.4k31533
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
asked Jan 12 at 20:14
Gabriel CorrêaGabriel Corrêa
143
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
add a comment |
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
add a comment |
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?
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
$begingroup$
Obrigado Lucas! Mas, devo fazer indução ?
$endgroup$
– Gabriel Corrêa
Jan 12 at 21:16
add a comment |
Your Answer
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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?
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
$begingroup$
Obrigado Lucas! Mas, devo fazer indução ?
$endgroup$
– Gabriel Corrêa
Jan 12 at 21:16
add a comment |
$begingroup$
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?
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
$begingroup$
Obrigado Lucas! Mas, devo fazer indução ?
$endgroup$
– Gabriel Corrêa
Jan 12 at 21:16
add a comment |
$begingroup$
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?
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
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?
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
answered Jan 12 at 20:44
Lucas CorrêaLucas Corrêa
1,5751421
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
add a comment |
$begingroup$
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
add a comment |
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1
$begingroup$
$x_n=y_n$.{}{}{}{}{}{}
$endgroup$
– hamam_Abdallah
Jan 12 at 20:18