Mixing
Altering separable space's definition
$begingroup$
A separable space has by definition a countable sub-space that is dense.
What if we replace "dense" by "its completion is dense"
What about "its completion is the whole space"? Do such spaces have names (if they are different)?
I guess my question only applies to metric spaces with the associated topology
general-topology separable-spaces
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
A separable space has by definition a countable sub-space that is dense.
What if we replace "dense" by "its completion is dense"
What about "its completion is the whole space"? Do such spaces have names (if they are different)?
I guess my question only applies to metric spaces with the associated topology
general-topology separable-spaces
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
$endgroup$
$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13
add a comment |
$begingroup$
A separable space has by definition a countable sub-space that is dense.
What if we replace "dense" by "its completion is dense"
What about "its completion is the whole space"? Do such spaces have names (if they are different)?
I guess my question only applies to metric spaces with the associated topology
general-topology separable-spaces
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
$endgroup$
A separable space has by definition a countable sub-space that is dense.
What if we replace "dense" by "its completion is dense"
What about "its completion is the whole space"? Do such spaces have names (if they are different)?
I guess my question only applies to metric spaces with the associated topology
general-topology separable-spaces
general-topology separable-spaces
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
edited Jan 21 at 10:16
John Cataldo
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 10:09
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13
add a comment |
$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13
$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
add a comment |
$begingroup$
(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.
(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.
For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $Esubset Dtimes Bbb Q^+,$ where $e=(delta,q)in Eiff B_d(delta,q)ne phi.$ Let $f:Eto Y$ be a function with $f(delta,q)in Ycap B_d(delta,q).$
Now $F={f(e):ein E}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $yin Usubset Y,$ where $U$ is open in $Y,$ then take $rin Bbb R^+$ such that $Ycap B_d(y,r)subset U.$
Now take $qin Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $delta in D cap B_d(y,q),$ so $yin B_d(delta,q)cap Y,$ so $(delta, q)in E.$ And by the triangle inequality (since $3q<r$ and $d(y,delta)<q $ ) we have $B_d(delta,q))subset B_d(y,r).$ So $Fcap U$ is not empty because $$f(delta,q)in Fcap B_d(delta,q)subset Fcap B_d(y,r)=$$ $$=(Fcap Y)cap B_d(y,r)=Fcap (Ycap B_d(y,r))subset Fcap U.$$
For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(,Cl_Y(G),)=Cl_X (,Ycap Cl_X(G),)subset Cl_X(,Cl_X(G),)=$$ $$=Cl_X(G)subset X.$$ So $X=Cl_X(G).$
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
add a comment |
$begingroup$
If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
add a comment |
$begingroup$
If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 10:18
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
65.3k42766
add a comment |
add a comment |
$begingroup$
(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.
(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.
For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $Esubset Dtimes Bbb Q^+,$ where $e=(delta,q)in Eiff B_d(delta,q)ne phi.$ Let $f:Eto Y$ be a function with $f(delta,q)in Ycap B_d(delta,q).$
Now $F={f(e):ein E}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $yin Usubset Y,$ where $U$ is open in $Y,$ then take $rin Bbb R^+$ such that $Ycap B_d(y,r)subset U.$
Now take $qin Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $delta in D cap B_d(y,q),$ so $yin B_d(delta,q)cap Y,$ so $(delta, q)in E.$ And by the triangle inequality (since $3q<r$ and $d(y,delta)<q $ ) we have $B_d(delta,q))subset B_d(y,r).$ So $Fcap U$ is not empty because $$f(delta,q)in Fcap B_d(delta,q)subset Fcap B_d(y,r)=$$ $$=(Fcap Y)cap B_d(y,r)=Fcap (Ycap B_d(y,r))subset Fcap U.$$
For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(,Cl_Y(G),)=Cl_X (,Ycap Cl_X(G),)subset Cl_X(,Cl_X(G),)=$$ $$=Cl_X(G)subset X.$$ So $X=Cl_X(G).$
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.
(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.
For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $Esubset Dtimes Bbb Q^+,$ where $e=(delta,q)in Eiff B_d(delta,q)ne phi.$ Let $f:Eto Y$ be a function with $f(delta,q)in Ycap B_d(delta,q).$
Now $F={f(e):ein E}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $yin Usubset Y,$ where $U$ is open in $Y,$ then take $rin Bbb R^+$ such that $Ycap B_d(y,r)subset U.$
Now take $qin Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $delta in D cap B_d(y,q),$ so $yin B_d(delta,q)cap Y,$ so $(delta, q)in E.$ And by the triangle inequality (since $3q<r$ and $d(y,delta)<q $ ) we have $B_d(delta,q))subset B_d(y,r).$ So $Fcap U$ is not empty because $$f(delta,q)in Fcap B_d(delta,q)subset Fcap B_d(y,r)=$$ $$=(Fcap Y)cap B_d(y,r)=Fcap (Ycap B_d(y,r))subset Fcap U.$$
For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(,Cl_Y(G),)=Cl_X (,Ycap Cl_X(G),)subset Cl_X(,Cl_X(G),)=$$ $$=Cl_X(G)subset X.$$ So $X=Cl_X(G).$
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.
(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.
For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $Esubset Dtimes Bbb Q^+,$ where $e=(delta,q)in Eiff B_d(delta,q)ne phi.$ Let $f:Eto Y$ be a function with $f(delta,q)in Ycap B_d(delta,q).$
Now $F={f(e):ein E}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $yin Usubset Y,$ where $U$ is open in $Y,$ then take $rin Bbb R^+$ such that $Ycap B_d(y,r)subset U.$
Now take $qin Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $delta in D cap B_d(y,q),$ so $yin B_d(delta,q)cap Y,$ so $(delta, q)in E.$ And by the triangle inequality (since $3q<r$ and $d(y,delta)<q $ ) we have $B_d(delta,q))subset B_d(y,r).$ So $Fcap U$ is not empty because $$f(delta,q)in Fcap B_d(delta,q)subset Fcap B_d(y,r)=$$ $$=(Fcap Y)cap B_d(y,r)=Fcap (Ycap B_d(y,r))subset Fcap U.$$
For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(,Cl_Y(G),)=Cl_X (,Ycap Cl_X(G),)subset Cl_X(,Cl_X(G),)=$$ $$=Cl_X(G)subset X.$$ So $X=Cl_X(G).$
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.
(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.
For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $Esubset Dtimes Bbb Q^+,$ where $e=(delta,q)in Eiff B_d(delta,q)ne phi.$ Let $f:Eto Y$ be a function with $f(delta,q)in Ycap B_d(delta,q).$
Now $F={f(e):ein E}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $yin Usubset Y,$ where $U$ is open in $Y,$ then take $rin Bbb R^+$ such that $Ycap B_d(y,r)subset U.$
Now take $qin Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $delta in D cap B_d(y,q),$ so $yin B_d(delta,q)cap Y,$ so $(delta, q)in E.$ And by the triangle inequality (since $3q<r$ and $d(y,delta)<q $ ) we have $B_d(delta,q))subset B_d(y,r).$ So $Fcap U$ is not empty because $$f(delta,q)in Fcap B_d(delta,q)subset Fcap B_d(y,r)=$$ $$=(Fcap Y)cap B_d(y,r)=Fcap (Ycap B_d(y,r))subset Fcap U.$$
For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(,Cl_Y(G),)=Cl_X (,Ycap Cl_X(G),)subset Cl_X(,Cl_X(G),)=$$ $$=Cl_X(G)subset X.$$ So $X=Cl_X(G).$
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
edited Jan 21 at 13:10
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 21 at 12:50
DanielWainfleetDanielWainfleet
35.4k31648
35.4k31648
add a comment |
add a comment |
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$begingroup$
How do you define completion in general topology?
$endgroup$
– Mindlack
Jan 21 at 10:12
$begingroup$
What do you mean by completion? Is it closure?
$endgroup$
– José Carlos Santos
Jan 21 at 10:13