Sequence of a dense set

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Let $mathbbX$ be a metric space with metric function $d_mathbbX(cdot,cdot):mathbbXtomathbbR_+$. Suppose $YsubseteqmathbbX$ is a dense set. Let $y_1,y_2,y_3,...subsetneq Y$ be a countable sequence. Is it true that this sequence is also dense in $mathbbX$? If not under what conditions would it be true?




real-analysis general-topology




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    In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
    – Anurag A
    Aug 3 at 18:50











  • @AnuragA, your comment should be an answer.
    – Chickenmancer
    Aug 3 at 19:13






  • 1




    In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
    – DanielWainfleet
    Aug 3 at 19:34















up vote
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down vote

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Let $mathbbX$ be a metric space with metric function $d_mathbbX(cdot,cdot):mathbbXtomathbbR_+$. Suppose $YsubseteqmathbbX$ is a dense set. Let $y_1,y_2,y_3,...subsetneq Y$ be a countable sequence. Is it true that this sequence is also dense in $mathbbX$? If not under what conditions would it be true?




real-analysis general-topology




share|cite|improve this question

















  • 8




    In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
    – Anurag A
    Aug 3 at 18:50











  • @AnuragA, your comment should be an answer.
    – Chickenmancer
    Aug 3 at 19:13






  • 1




    In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
    – DanielWainfleet
    Aug 3 at 19:34













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
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1





Let $mathbbX$ be a metric space with metric function $d_mathbbX(cdot,cdot):mathbbXtomathbbR_+$. Suppose $YsubseteqmathbbX$ is a dense set. Let $y_1,y_2,y_3,...subsetneq Y$ be a countable sequence. Is it true that this sequence is also dense in $mathbbX$? If not under what conditions would it be true?




real-analysis general-topology




share|cite|improve this question













Let $mathbbX$ be a metric space with metric function $d_mathbbX(cdot,cdot):mathbbXtomathbbR_+$. Suppose $YsubseteqmathbbX$ is a dense set. Let $y_1,y_2,y_3,...subsetneq Y$ be a countable sequence. Is it true that this sequence is also dense in $mathbbX$? If not under what conditions would it be true?





real-analysis general-topology





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 19:11









Andrés E. Caicedo

63k7151235




63k7151235









asked Aug 3 at 18:47









Arian

5,235817




5,235817







  • 8




    In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
    – Anurag A
    Aug 3 at 18:50











  • @AnuragA, your comment should be an answer.
    – Chickenmancer
    Aug 3 at 19:13






  • 1




    In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
    – DanielWainfleet
    Aug 3 at 19:34













  • 8




    In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
    – Anurag A
    Aug 3 at 18:50











  • @AnuragA, your comment should be an answer.
    – Chickenmancer
    Aug 3 at 19:13






  • 1




    In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
    – DanielWainfleet
    Aug 3 at 19:34








8




8




In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
– Anurag A
Aug 3 at 18:50





In general the statement is not true. Just take with the standard metric, $X=mathbbR, Y=mathbbQ$ and the countable subset as $mathbbZ$.
– Anurag A
Aug 3 at 18:50













@AnuragA, your comment should be an answer.
– Chickenmancer
Aug 3 at 19:13




@AnuragA, your comment should be an answer.
– Chickenmancer
Aug 3 at 19:13




1




1




In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
– DanielWainfleet
Aug 3 at 19:34





In general a metric space, or any topological space, need not have a countable dense subset. For example let $Bbb X=Bbb R$ (or any other uncountable set) and let $d(a,b)=1$ whenever $a ne b$ ( "the" discrete metric). Then the only dense subset of $Bbb X$ is $Bbb X$.
– DanielWainfleet
Aug 3 at 19:34
















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