Difference in the definitions of density in topology?

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I have been trying to understand the concept of density and have so far come across two definitions. That is, the set $A$ is dense in a space $B$ if



cl(A) = B,



however, I have also come across the following definition.



The set $A$ is dense in $B$ if



$A subset B$ and $B subseteq cl(A)$.



Just wondering if the two definitions are indeed the same.



Also, additionally, we have it given that



$B subseteq cl(A)$.



Taking the closure of both sides yields



$cl(B) subseteq cl(A)$.



I am also just wondering if this is also indeed true.



Thanks for the help.










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  • The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
    – Fabio Somenzi
    Sep 1 at 3:39










  • For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
    – Fabio Somenzi
    Sep 1 at 3:43











  • Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
    – JasKa071
    Sep 1 at 4:27










  • Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
    – Fabio Somenzi
    Sep 1 at 4:57















up vote
0
down vote

favorite












I have been trying to understand the concept of density and have so far come across two definitions. That is, the set $A$ is dense in a space $B$ if



cl(A) = B,



however, I have also come across the following definition.



The set $A$ is dense in $B$ if



$A subset B$ and $B subseteq cl(A)$.



Just wondering if the two definitions are indeed the same.



Also, additionally, we have it given that



$B subseteq cl(A)$.



Taking the closure of both sides yields



$cl(B) subseteq cl(A)$.



I am also just wondering if this is also indeed true.



Thanks for the help.










share|cite|improve this question





















  • The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
    – Fabio Somenzi
    Sep 1 at 3:39










  • For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
    – Fabio Somenzi
    Sep 1 at 3:43











  • Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
    – JasKa071
    Sep 1 at 4:27










  • Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
    – Fabio Somenzi
    Sep 1 at 4:57













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have been trying to understand the concept of density and have so far come across two definitions. That is, the set $A$ is dense in a space $B$ if



cl(A) = B,



however, I have also come across the following definition.



The set $A$ is dense in $B$ if



$A subset B$ and $B subseteq cl(A)$.



Just wondering if the two definitions are indeed the same.



Also, additionally, we have it given that



$B subseteq cl(A)$.



Taking the closure of both sides yields



$cl(B) subseteq cl(A)$.



I am also just wondering if this is also indeed true.



Thanks for the help.










share|cite|improve this question













I have been trying to understand the concept of density and have so far come across two definitions. That is, the set $A$ is dense in a space $B$ if



cl(A) = B,



however, I have also come across the following definition.



The set $A$ is dense in $B$ if



$A subset B$ and $B subseteq cl(A)$.



Just wondering if the two definitions are indeed the same.



Also, additionally, we have it given that



$B subseteq cl(A)$.



Taking the closure of both sides yields



$cl(B) subseteq cl(A)$.



I am also just wondering if this is also indeed true.



Thanks for the help.







general-topology metric-spaces






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asked Sep 1 at 3:17









JasKa071

112




112











  • The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
    – Fabio Somenzi
    Sep 1 at 3:39










  • For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
    – Fabio Somenzi
    Sep 1 at 3:43











  • Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
    – JasKa071
    Sep 1 at 4:27










  • Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
    – Fabio Somenzi
    Sep 1 at 4:57

















  • The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
    – Fabio Somenzi
    Sep 1 at 3:39










  • For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
    – Fabio Somenzi
    Sep 1 at 3:43











  • Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
    – JasKa071
    Sep 1 at 4:27










  • Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
    – Fabio Somenzi
    Sep 1 at 4:57
















The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
– Fabio Somenzi
Sep 1 at 3:39




The first definition, in which $B$ is the topological space, is a special case of the second definition, in which $B$ is any subset of the topological space such that $A subset B$.
– Fabio Somenzi
Sep 1 at 3:39












For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
– Fabio Somenzi
Sep 1 at 3:43





For the second part, note that $A subseteq B$ implies $cl(A) subseteq cl(B)$ and $cl(cl(A)) = cl(A)$.
– Fabio Somenzi
Sep 1 at 3:43













Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
– JasKa071
Sep 1 at 4:27




Just wondering if it holds the other way as well, that is, $cl(B) subseteq cl(A)$?
– JasKa071
Sep 1 at 4:27












Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
– Fabio Somenzi
Sep 1 at 4:57





Apply "$X subseteq Y$ implies $cl(X) subseteq cl(Y)$" to $B$ and $cl(A)$.
– Fabio Somenzi
Sep 1 at 4:57











1 Answer
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So these are slightly different notions linguistically:



the first is the more standard one and is for spaces $X$: $A$ (a subset of $X$) is called
dense in $X$ when $operatornamecl(A) = X$.



The second notion is for subsets of a space $X$ : the set $A$ is called dense in $B$ whenever $A subseteq B$ and $B subseteq operatornamecl(B)$.



But this second notion is not really different in content: it means that $operatornamecl_B(A) = A$, where $operatornamecl_B$ is the closure operator in $B$ with the subspace topology seen as a space in its own right.



This can be seen by applying the following general fact for subspaces and their closure operator, and how it relates to the closure in the ambient space:




Fact: If $B$ is a subset of $X$ in the subspace topology w.r.t. $X$, and $A subseteq B$ then $operatornamecl_B(A) = operatornamecl(A) cap B$; or in words: the closure of $A$ in the subspace $B$ is just the part of the full closure of $A$ (in $X$) that lies inside $B$.




This fact can be proved directly from the definition of the subspace topology and the characterisation of the closure of $A$ as all points $x$ such that all neighbourhoods of $x$ intersect $A$, or from the fact that closed sets of $B$ are closed sets of $X$ intersected with $B$, and the closure of $A$ is the smallest closed set that contains $A$.



Now suppose $A subseteq B$ and $A$ is dense (in the first sense) in the subspace $B$, so $operatornamecl_B(A) = B$ and by the fact we can rewrite this as



$$operatornamecl_B(A) = operatornamecl(A) cap B = B$$ but simple set theory tells us that $$C cap B = B text iff B subseteq C$$ for all $B,C$ and so we actually get that



$$operatornamecl_B(A) = B text iff B subseteq operatornamecl(A)$$



So the "notions" of dense mean the same thing: $A$ is dense in $B$ if we can "approximate all points of $B$ by points of $A$", or $B$ is a subset of $operatornamecl(A)$ (where $operatornamecl(A)$ can be thought of as "all points that can be approximated by points of $A$", intuitively (or more formally e.g. if you use filters or nets)






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    1 Answer
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    oldest

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    active

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    up vote
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    down vote













    So these are slightly different notions linguistically:



    the first is the more standard one and is for spaces $X$: $A$ (a subset of $X$) is called
    dense in $X$ when $operatornamecl(A) = X$.



    The second notion is for subsets of a space $X$ : the set $A$ is called dense in $B$ whenever $A subseteq B$ and $B subseteq operatornamecl(B)$.



    But this second notion is not really different in content: it means that $operatornamecl_B(A) = A$, where $operatornamecl_B$ is the closure operator in $B$ with the subspace topology seen as a space in its own right.



    This can be seen by applying the following general fact for subspaces and their closure operator, and how it relates to the closure in the ambient space:




    Fact: If $B$ is a subset of $X$ in the subspace topology w.r.t. $X$, and $A subseteq B$ then $operatornamecl_B(A) = operatornamecl(A) cap B$; or in words: the closure of $A$ in the subspace $B$ is just the part of the full closure of $A$ (in $X$) that lies inside $B$.




    This fact can be proved directly from the definition of the subspace topology and the characterisation of the closure of $A$ as all points $x$ such that all neighbourhoods of $x$ intersect $A$, or from the fact that closed sets of $B$ are closed sets of $X$ intersected with $B$, and the closure of $A$ is the smallest closed set that contains $A$.



    Now suppose $A subseteq B$ and $A$ is dense (in the first sense) in the subspace $B$, so $operatornamecl_B(A) = B$ and by the fact we can rewrite this as



    $$operatornamecl_B(A) = operatornamecl(A) cap B = B$$ but simple set theory tells us that $$C cap B = B text iff B subseteq C$$ for all $B,C$ and so we actually get that



    $$operatornamecl_B(A) = B text iff B subseteq operatornamecl(A)$$



    So the "notions" of dense mean the same thing: $A$ is dense in $B$ if we can "approximate all points of $B$ by points of $A$", or $B$ is a subset of $operatornamecl(A)$ (where $operatornamecl(A)$ can be thought of as "all points that can be approximated by points of $A$", intuitively (or more formally e.g. if you use filters or nets)






    share|cite|improve this answer
























      up vote
      0
      down vote













      So these are slightly different notions linguistically:



      the first is the more standard one and is for spaces $X$: $A$ (a subset of $X$) is called
      dense in $X$ when $operatornamecl(A) = X$.



      The second notion is for subsets of a space $X$ : the set $A$ is called dense in $B$ whenever $A subseteq B$ and $B subseteq operatornamecl(B)$.



      But this second notion is not really different in content: it means that $operatornamecl_B(A) = A$, where $operatornamecl_B$ is the closure operator in $B$ with the subspace topology seen as a space in its own right.



      This can be seen by applying the following general fact for subspaces and their closure operator, and how it relates to the closure in the ambient space:




      Fact: If $B$ is a subset of $X$ in the subspace topology w.r.t. $X$, and $A subseteq B$ then $operatornamecl_B(A) = operatornamecl(A) cap B$; or in words: the closure of $A$ in the subspace $B$ is just the part of the full closure of $A$ (in $X$) that lies inside $B$.




      This fact can be proved directly from the definition of the subspace topology and the characterisation of the closure of $A$ as all points $x$ such that all neighbourhoods of $x$ intersect $A$, or from the fact that closed sets of $B$ are closed sets of $X$ intersected with $B$, and the closure of $A$ is the smallest closed set that contains $A$.



      Now suppose $A subseteq B$ and $A$ is dense (in the first sense) in the subspace $B$, so $operatornamecl_B(A) = B$ and by the fact we can rewrite this as



      $$operatornamecl_B(A) = operatornamecl(A) cap B = B$$ but simple set theory tells us that $$C cap B = B text iff B subseteq C$$ for all $B,C$ and so we actually get that



      $$operatornamecl_B(A) = B text iff B subseteq operatornamecl(A)$$



      So the "notions" of dense mean the same thing: $A$ is dense in $B$ if we can "approximate all points of $B$ by points of $A$", or $B$ is a subset of $operatornamecl(A)$ (where $operatornamecl(A)$ can be thought of as "all points that can be approximated by points of $A$", intuitively (or more formally e.g. if you use filters or nets)






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        So these are slightly different notions linguistically:



        the first is the more standard one and is for spaces $X$: $A$ (a subset of $X$) is called
        dense in $X$ when $operatornamecl(A) = X$.



        The second notion is for subsets of a space $X$ : the set $A$ is called dense in $B$ whenever $A subseteq B$ and $B subseteq operatornamecl(B)$.



        But this second notion is not really different in content: it means that $operatornamecl_B(A) = A$, where $operatornamecl_B$ is the closure operator in $B$ with the subspace topology seen as a space in its own right.



        This can be seen by applying the following general fact for subspaces and their closure operator, and how it relates to the closure in the ambient space:




        Fact: If $B$ is a subset of $X$ in the subspace topology w.r.t. $X$, and $A subseteq B$ then $operatornamecl_B(A) = operatornamecl(A) cap B$; or in words: the closure of $A$ in the subspace $B$ is just the part of the full closure of $A$ (in $X$) that lies inside $B$.




        This fact can be proved directly from the definition of the subspace topology and the characterisation of the closure of $A$ as all points $x$ such that all neighbourhoods of $x$ intersect $A$, or from the fact that closed sets of $B$ are closed sets of $X$ intersected with $B$, and the closure of $A$ is the smallest closed set that contains $A$.



        Now suppose $A subseteq B$ and $A$ is dense (in the first sense) in the subspace $B$, so $operatornamecl_B(A) = B$ and by the fact we can rewrite this as



        $$operatornamecl_B(A) = operatornamecl(A) cap B = B$$ but simple set theory tells us that $$C cap B = B text iff B subseteq C$$ for all $B,C$ and so we actually get that



        $$operatornamecl_B(A) = B text iff B subseteq operatornamecl(A)$$



        So the "notions" of dense mean the same thing: $A$ is dense in $B$ if we can "approximate all points of $B$ by points of $A$", or $B$ is a subset of $operatornamecl(A)$ (where $operatornamecl(A)$ can be thought of as "all points that can be approximated by points of $A$", intuitively (or more formally e.g. if you use filters or nets)






        share|cite|improve this answer












        So these are slightly different notions linguistically:



        the first is the more standard one and is for spaces $X$: $A$ (a subset of $X$) is called
        dense in $X$ when $operatornamecl(A) = X$.



        The second notion is for subsets of a space $X$ : the set $A$ is called dense in $B$ whenever $A subseteq B$ and $B subseteq operatornamecl(B)$.



        But this second notion is not really different in content: it means that $operatornamecl_B(A) = A$, where $operatornamecl_B$ is the closure operator in $B$ with the subspace topology seen as a space in its own right.



        This can be seen by applying the following general fact for subspaces and their closure operator, and how it relates to the closure in the ambient space:




        Fact: If $B$ is a subset of $X$ in the subspace topology w.r.t. $X$, and $A subseteq B$ then $operatornamecl_B(A) = operatornamecl(A) cap B$; or in words: the closure of $A$ in the subspace $B$ is just the part of the full closure of $A$ (in $X$) that lies inside $B$.




        This fact can be proved directly from the definition of the subspace topology and the characterisation of the closure of $A$ as all points $x$ such that all neighbourhoods of $x$ intersect $A$, or from the fact that closed sets of $B$ are closed sets of $X$ intersected with $B$, and the closure of $A$ is the smallest closed set that contains $A$.



        Now suppose $A subseteq B$ and $A$ is dense (in the first sense) in the subspace $B$, so $operatornamecl_B(A) = B$ and by the fact we can rewrite this as



        $$operatornamecl_B(A) = operatornamecl(A) cap B = B$$ but simple set theory tells us that $$C cap B = B text iff B subseteq C$$ for all $B,C$ and so we actually get that



        $$operatornamecl_B(A) = B text iff B subseteq operatornamecl(A)$$



        So the "notions" of dense mean the same thing: $A$ is dense in $B$ if we can "approximate all points of $B$ by points of $A$", or $B$ is a subset of $operatornamecl(A)$ (where $operatornamecl(A)$ can be thought of as "all points that can be approximated by points of $A$", intuitively (or more formally e.g. if you use filters or nets)







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        answered Sep 1 at 5:38









        Henno Brandsma

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