Banach space with subset whose elements are at least $dgt 0$ far from each other is not separable












1















Let $X$ be a Banach space, and $Asubseteq X$ subgroup, where $A$ is not countable, and there is some $d gt 0$ such that for all $x,y in A$: $||x-y||>d$. Prove that $X$ is not separable.




My attempt -



Let's assume that $X$ is separable, and let $B$ be the dense countable subset of $X$. Than $Asubseteq overline B$. Meaning that every element of $A$, is a limit point of some sequence in $B$.



So if I could find a Cauchy sequence in $B$, such that it's limit is in $A$ (that Cauchy sequence has a limit since $X$ is Banach), the property of elements in $A$(that every two elements are at least of distance $d$), would contradict the propeties of Cauchy sequences, and hence $X$ cannot be separable.



Two things that I havn't used are that $B$ is countable and that $A$ is not countable. How can I prove that there is at least one Cauchy sequence in $B$ that converges to some point in $A$?



In fact, how can I know at all that there is Cauchy sequence in $B$?



Or perhaps, is there any other way?






banach-spaces normed-spaces cauchy-sequences separable-spaces







share|cite|improve this question
























  • Any help? I'm really stuck here...
    – ChikChak
    Nov 24 at 17:39
















1















Let $X$ be a Banach space, and $Asubseteq X$ subgroup, where $A$ is not countable, and there is some $d gt 0$ such that for all $x,y in A$: $||x-y||>d$. Prove that $X$ is not separable.




My attempt -



Let's assume that $X$ is separable, and let $B$ be the dense countable subset of $X$. Than $Asubseteq overline B$. Meaning that every element of $A$, is a limit point of some sequence in $B$.



So if I could find a Cauchy sequence in $B$, such that it's limit is in $A$ (that Cauchy sequence has a limit since $X$ is Banach), the property of elements in $A$(that every two elements are at least of distance $d$), would contradict the propeties of Cauchy sequences, and hence $X$ cannot be separable.



Two things that I havn't used are that $B$ is countable and that $A$ is not countable. How can I prove that there is at least one Cauchy sequence in $B$ that converges to some point in $A$?



In fact, how can I know at all that there is Cauchy sequence in $B$?



Or perhaps, is there any other way?






banach-spaces normed-spaces cauchy-sequences separable-spaces







share|cite|improve this question
























  • Any help? I'm really stuck here...
    – ChikChak
    Nov 24 at 17:39














1












1








1








Let $X$ be a Banach space, and $Asubseteq X$ subgroup, where $A$ is not countable, and there is some $d gt 0$ such that for all $x,y in A$: $||x-y||>d$. Prove that $X$ is not separable.




My attempt -



Let's assume that $X$ is separable, and let $B$ be the dense countable subset of $X$. Than $Asubseteq overline B$. Meaning that every element of $A$, is a limit point of some sequence in $B$.



So if I could find a Cauchy sequence in $B$, such that it's limit is in $A$ (that Cauchy sequence has a limit since $X$ is Banach), the property of elements in $A$(that every two elements are at least of distance $d$), would contradict the propeties of Cauchy sequences, and hence $X$ cannot be separable.



Two things that I havn't used are that $B$ is countable and that $A$ is not countable. How can I prove that there is at least one Cauchy sequence in $B$ that converges to some point in $A$?



In fact, how can I know at all that there is Cauchy sequence in $B$?



Or perhaps, is there any other way?






banach-spaces normed-spaces cauchy-sequences separable-spaces







share|cite|improve this question
















Let $X$ be a Banach space, and $Asubseteq X$ subgroup, where $A$ is not countable, and there is some $d gt 0$ such that for all $x,y in A$: $||x-y||>d$. Prove that $X$ is not separable.




My attempt -



Let's assume that $X$ is separable, and let $B$ be the dense countable subset of $X$. Than $Asubseteq overline B$. Meaning that every element of $A$, is a limit point of some sequence in $B$.



So if I could find a Cauchy sequence in $B$, such that it's limit is in $A$ (that Cauchy sequence has a limit since $X$ is Banach), the property of elements in $A$(that every two elements are at least of distance $d$), would contradict the propeties of Cauchy sequences, and hence $X$ cannot be separable.



Two things that I havn't used are that $B$ is countable and that $A$ is not countable. How can I prove that there is at least one Cauchy sequence in $B$ that converges to some point in $A$?



In fact, how can I know at all that there is Cauchy sequence in $B$?



Or perhaps, is there any other way?






banach-spaces normed-spaces cauchy-sequences separable-spaces




banach-spaces normed-spaces cauchy-sequences separable-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 14:37









Ethan Bolker

40.9k546108




40.9k546108










asked Nov 23 at 17:04









ChikChak

800418




800418












  • Any help? I'm really stuck here...
    – ChikChak
    Nov 24 at 17:39


















  • Any help? I'm really stuck here...
    – ChikChak
    Nov 24 at 17:39
















Any help? I'm really stuck here...
– ChikChak
Nov 24 at 17:39




Any help? I'm really stuck here...
– ChikChak
Nov 24 at 17:39










1 Answer
1






active

oldest

votes


















1














$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a in A}$ given by
$$
U_a = B_a(d/3),
$$
Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.






share|cite|improve this answer





















  • Nice. So the assumption that $X$ is Banach is not necessary?
    – ChikChak
    Nov 25 at 14:32






  • 1




    I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
    – Adrián González-Pérez
    Nov 25 at 14:36











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1 Answer
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active

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1 Answer
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active

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1














$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a in A}$ given by
$$
U_a = B_a(d/3),
$$
Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.






share|cite|improve this answer





















  • Nice. So the assumption that $X$ is Banach is not necessary?
    – ChikChak
    Nov 25 at 14:32






  • 1




    I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
    – Adrián González-Pérez
    Nov 25 at 14:36
















1














$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a in A}$ given by
$$
U_a = B_a(d/3),
$$
Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.






share|cite|improve this answer





















  • Nice. So the assumption that $X$ is Banach is not necessary?
    – ChikChak
    Nov 25 at 14:32






  • 1




    I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
    – Adrián González-Pérez
    Nov 25 at 14:36














1












1








1






$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a in A}$ given by
$$
U_a = B_a(d/3),
$$
Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.






share|cite|improve this answer












$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a in A}$ given by
$$
U_a = B_a(d/3),
$$
Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 14:30









Adrián González-Pérez

926138




926138












  • Nice. So the assumption that $X$ is Banach is not necessary?
    – ChikChak
    Nov 25 at 14:32






  • 1




    I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
    – Adrián González-Pérez
    Nov 25 at 14:36


















  • Nice. So the assumption that $X$ is Banach is not necessary?
    – ChikChak
    Nov 25 at 14:32






  • 1




    I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
    – Adrián González-Pérez
    Nov 25 at 14:36
















Nice. So the assumption that $X$ is Banach is not necessary?
– ChikChak
Nov 25 at 14:32




Nice. So the assumption that $X$ is Banach is not necessary?
– ChikChak
Nov 25 at 14:32




1




1




I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
– Adrián González-Pérez
Nov 25 at 14:36




I think this works for general topological spaces. If your space admits an uncountable family of nonempty open sets it can not be separable. No need if metrics or completion.
– Adrián González-Pérez
Nov 25 at 14:36


















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