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 .
general-topology metric-spaces
edited Jan 10 at 10:49
bof
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
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add a comment |
$begingroup$
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 .
general-topology metric-spaces
edited Jan 10 at 10:49
bof
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
$endgroup$
add a comment |
$begingroup$
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 .
general-topology metric-spaces
edited Jan 10 at 10:49
bof
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
$endgroup$
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
general-topology metric-spaces
edited Jan 10 at 10:49
bof
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
edited Jan 10 at 10:49
bof
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
edited Jan 10 at 10:49
bof
52.6k559121
edited Jan 10 at 10:49
bof
52.6k559121
edited Jan 10 at 10:49
bof
52.6k559121
52.6k559121
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
asked Jan 10 at 10:34
RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY
565
565
add a comment |
add a comment |
1 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 $D$.
answered Jan 10 at 10:44
bofbof
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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 $D$.
answered Jan 10 at 10:44
bofbof
52.6k559121
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add a comment |
$begingroup$
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 $D$.
answered Jan 10 at 10:44
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
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 $D$.
answered Jan 10 at 10:44
bofbof
52.6k559121
$endgroup$
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 $D$.
answered Jan 10 at 10:44
bofbof
52.6k559121
answered Jan 10 at 10:44
bofbof
52.6k559121
answered Jan 10 at 10:44
bofbof
52.6k559121
answered Jan 10 at 10:44
bofbof
52.6k559121
52.6k559121
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