Let S be an uncountable subset of a separable metric space (X,d). Then,can we find any such S in X which...












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This question arrived into my mind after observing the fact that many of the uncountable subsets is either dense or posseses a dense subset, dense in itself, even the Cantor set in R is perfect .










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    This question arrived into my mind after observing the fact that many of the uncountable subsets is either dense or posseses a dense subset, dense in itself, even the Cantor set in R is perfect .










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      $begingroup$


      This question arrived into my mind after observing the fact that many of the uncountable subsets is either dense or posseses a dense subset, dense in itself, even the Cantor set in R is perfect .










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      This question arrived into my mind after observing the fact that many of the uncountable subsets is either dense or posseses a dense subset, dense in itself, even the Cantor set in R is perfect .







      general-topology metric-spaces






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      edited Jan 10 at 10:49









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      asked Jan 10 at 10:34









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          Since a subspace of a separable metric space is separable, we may as well assume that $S=X$. The answer is yes, any uncountable separable metric space $S$ contains an uncountable subset which is dense in itself. To see this write $S=Ccup D$ where $xin C$ if some neighborhood of $x$ is countable, $xin D$ if every neighborhood of $x$ is uncountable. Now it is easy to see that $C$ is countable (from the Lindelöf property of separable metric spaces); it follows that $D$ is uncountable, and every neighborhood of a point $xin D$ contains uncountably many points of $D$.






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            $begingroup$

            Since a subspace of a separable metric space is separable, we may as well assume that $S=X$. The answer is yes, any uncountable separable metric space $S$ contains an uncountable subset which is dense in itself. To see this write $S=Ccup D$ where $xin C$ if some neighborhood of $x$ is countable, $xin D$ if every neighborhood of $x$ is uncountable. Now it is easy to see that $C$ is countable (from the Lindelöf property of separable metric spaces); it follows that $D$ is uncountable, and every neighborhood of a point $xin D$ contains uncountably many points of $D$.






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              $begingroup$

              Since a subspace of a separable metric space is separable, we may as well assume that $S=X$. The answer is yes, any uncountable separable metric space $S$ contains an uncountable subset which is dense in itself. To see this write $S=Ccup D$ where $xin C$ if some neighborhood of $x$ is countable, $xin D$ if every neighborhood of $x$ is uncountable. Now it is easy to see that $C$ is countable (from the Lindelöf property of separable metric spaces); it follows that $D$ is uncountable, and every neighborhood of a point $xin D$ contains uncountably many points of $D$.






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                $begingroup$

                Since a subspace of a separable metric space is separable, we may as well assume that $S=X$. The answer is yes, any uncountable separable metric space $S$ contains an uncountable subset which is dense in itself. To see this write $S=Ccup D$ where $xin C$ if some neighborhood of $x$ is countable, $xin D$ if every neighborhood of $x$ is uncountable. Now it is easy to see that $C$ is countable (from the Lindelöf property of separable metric spaces); it follows that $D$ is uncountable, and every neighborhood of a point $xin D$ contains uncountably many points of $D$.






                share|cite|improve this answer









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                Since a subspace of a separable metric space is separable, we may as well assume that $S=X$. The answer is yes, any uncountable separable metric space $S$ contains an uncountable subset which is dense in itself. To see this write $S=Ccup D$ where $xin C$ if some neighborhood of $x$ is countable, $xin D$ if every neighborhood of $x$ is uncountable. Now it is easy to see that $C$ is countable (from the Lindelöf property of separable metric spaces); it follows that $D$ is uncountable, and every neighborhood of a point $xin D$ contains uncountably many points of $D$.







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                answered Jan 10 at 10:44









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