Banach space with subset whose elements are at least $dgt 0$ far from each other is not separable
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
add a comment |
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
Any help? I'm really stuck here...
– ChikChak
Nov 24 at 17:39
add a comment |
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
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
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
add a comment |
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
add a comment |
1 Answer
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$X$ 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.
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
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$X$ 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.
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
add a comment |
$X$ 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.
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
add a comment |
$X$ 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.
$X$ 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.
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
add a comment |
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
add a comment |
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Any help? I'm really stuck here...
– ChikChak
Nov 24 at 17:39