Metric spaces and compactness
$begingroup$
Let $X$ be a metric space and $Y$ a subset of $X$. Suppose that from every sequence in $Y$ it is possible to extract a subsequence that converges to some point of $X$. I want to show that the closure of $Y$ is compact. Let $(x_k)$ be a sequence in $overline{Y}$. I want to show that it admits a convergent subsequence (obviously if this is the case, the limit will be automatically a poin in $overline{Y})$. Any suggestion? Thank-you.
real-analysis calculus general-topology
asked Dec 23 '18 at 17:02
MinatoMinato
549313
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add a comment |
$begingroup$
Let $X$ be a metric space and $Y$ a subset of $X$. Suppose that from every sequence in $Y$ it is possible to extract a subsequence that converges to some point of $X$. I want to show that the closure of $Y$ is compact. Let $(x_k)$ be a sequence in $overline{Y}$. I want to show that it admits a convergent subsequence (obviously if this is the case, the limit will be automatically a poin in $overline{Y})$. Any suggestion? Thank-you.
real-analysis calculus general-topology
asked Dec 23 '18 at 17:02
MinatoMinato
549313
$endgroup$
add a comment |
$begingroup$
Let $X$ be a metric space and $Y$ a subset of $X$. Suppose that from every sequence in $Y$ it is possible to extract a subsequence that converges to some point of $X$. I want to show that the closure of $Y$ is compact. Let $(x_k)$ be a sequence in $overline{Y}$. I want to show that it admits a convergent subsequence (obviously if this is the case, the limit will be automatically a poin in $overline{Y})$. Any suggestion? Thank-you.
real-analysis calculus general-topology
asked Dec 23 '18 at 17:02
MinatoMinato
549313
$endgroup$
Let $X$ be a metric space and $Y$ a subset of $X$. Suppose that from every sequence in $Y$ it is possible to extract a subsequence that converges to some point of $X$. I want to show that the closure of $Y$ is compact. Let $(x_k)$ be a sequence in $overline{Y}$. I want to show that it admits a convergent subsequence (obviously if this is the case, the limit will be automatically a poin in $overline{Y})$. Any suggestion? Thank-you.
real-analysis calculus general-topology
real-analysis calculus general-topology
asked Dec 23 '18 at 17:02
MinatoMinato
549313
asked Dec 23 '18 at 17:02
MinatoMinato
549313
asked Dec 23 '18 at 17:02
MinatoMinato
549313
asked Dec 23 '18 at 17:02
MinatoMinato
549313
asked Dec 23 '18 at 17:02
MinatoMinato
549313
549313
add a comment |
add a comment |
1 Answer
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Let $x_k$ be a sequence of $bar Y$, consider $y_kin Y$ with $d(x_k,y_k)<{1over 2^k}$, you can extract a subsequence $y_{n_k}$ from $y_k$, that converges towards $y$. $d(y,x_{n_k})leq d(y,y_{n_k})+d(y_{n_k},x_{n_k})$. This implies that $x_{n_k}$ converges towards $y$.
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
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1 Answer
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1 Answer
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$begingroup$
Let $x_k$ be a sequence of $bar Y$, consider $y_kin Y$ with $d(x_k,y_k)<{1over 2^k}$, you can extract a subsequence $y_{n_k}$ from $y_k$, that converges towards $y$. $d(y,x_{n_k})leq d(y,y_{n_k})+d(y_{n_k},x_{n_k})$. This implies that $x_{n_k}$ converges towards $y$.
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
add a comment |
$begingroup$
Let $x_k$ be a sequence of $bar Y$, consider $y_kin Y$ with $d(x_k,y_k)<{1over 2^k}$, you can extract a subsequence $y_{n_k}$ from $y_k$, that converges towards $y$. $d(y,x_{n_k})leq d(y,y_{n_k})+d(y_{n_k},x_{n_k})$. This implies that $x_{n_k}$ converges towards $y$.
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
add a comment |
$begingroup$
Let $x_k$ be a sequence of $bar Y$, consider $y_kin Y$ with $d(x_k,y_k)<{1over 2^k}$, you can extract a subsequence $y_{n_k}$ from $y_k$, that converges towards $y$. $d(y,x_{n_k})leq d(y,y_{n_k})+d(y_{n_k},x_{n_k})$. This implies that $x_{n_k}$ converges towards $y$.
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
$endgroup$
Let $x_k$ be a sequence of $bar Y$, consider $y_kin Y$ with $d(x_k,y_k)<{1over 2^k}$, you can extract a subsequence $y_{n_k}$ from $y_k$, that converges towards $y$. $d(y,x_{n_k})leq d(y,y_{n_k})+d(y_{n_k},x_{n_k})$. This implies that $x_{n_k}$ converges towards $y$.
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
edited Dec 23 '18 at 17:15
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
answered Dec 23 '18 at 17:06
Tsemo AristideTsemo Aristide
59.2k11445
59.2k11445
add a comment |
add a comment |
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