Why does $A_j subseteq cl ( bigcup_m=1^n A_m )$ imply $cl(A_j) subseteq cl ( bigcup_m=1^n A_m )$? The 2019 Stack Overflow Developer Survey Results Are InProve $bigcup _j=1^nA_j subseteq bigcup_j = 1^n B_j$Inner and outer limit of sets sequences$sigma(mathcalA) = $ the set of countable unions of countable intersections of elements or complements of elements of $mathcalA$$if An subseteq A $ for all $n in mathbbN, $ then $ bigcup_n=1^infty An subseteq A $Why does $x in A implies x in B$ imply $A subset B$ and not $B subset A$?Why is it the case that $mgeq n$, we have that $bigcap_k = n^infty A_k subseteq A_m$?Why does this not show $P(Acup B) subseteq P(A)cup P(B)$?Does $Ain B$ imply $Asubset B$?For each of the following formulae ϕ(x), show why the set x : ϕ(x) does or does not exist:Countable union of disjoint sets
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Why does $A_j subseteq cl ( bigcup_m=1^n A_m )$ imply $cl(A_j) subseteq cl ( bigcup_m=1^n A_m )$?
The 2019 Stack Overflow Developer Survey Results Are InProve $bigcup _j=1^nA_j subseteq bigcup_j = 1^n B_j$Inner and outer limit of sets sequences$sigma(mathcalA) = $ the set of countable unions of countable intersections of elements or complements of elements of $mathcalA$$if An subseteq A $ for all $n in mathbbN, $ then $ bigcup_n=1^infty An subseteq A $Why does $x in A implies x in B$ imply $A subset B$ and not $B subset A$?Why is it the case that $mgeq n$, we have that $bigcap_k = n^infty A_k subseteq A_m$?Why does this not show $P(Acup B) subseteq P(A)cup P(B)$?Does $Ain B$ imply $Asubset B$?For each of the following formulae ϕ(x), show why the set x : ϕ(x) does or does not exist:Countable union of disjoint sets
1
$begingroup$
Why does $A_j subseteq cl bigg( cup_m=1^n A_m bigg)$ imply $cl(A_j) subseteq cl bigg( cup_m=1^n A_m bigg)$?
This is intuitive, but I was thinking of whether one can be sure that there are no counterexamples.
elementary-set-theory
edited Mar 23 at 18:04
Henning Makholm
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
$endgroup$
add a comment |
1
$begingroup$
Why does $A_j subseteq cl bigg( cup_m=1^n A_m bigg)$ imply $cl(A_j) subseteq cl bigg( cup_m=1^n A_m bigg)$?
This is intuitive, but I was thinking of whether one can be sure that there are no counterexamples.
elementary-set-theory
edited Mar 23 at 18:04
Henning Makholm
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
$endgroup$
$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06
add a comment |
1
1
1
$begingroup$
Why does $A_j subseteq cl bigg( cup_m=1^n A_m bigg)$ imply $cl(A_j) subseteq cl bigg( cup_m=1^n A_m bigg)$?
This is intuitive, but I was thinking of whether one can be sure that there are no counterexamples.
elementary-set-theory
edited Mar 23 at 18:04
Henning Makholm
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
$endgroup$
Why does $A_j subseteq cl bigg( cup_m=1^n A_m bigg)$ imply $cl(A_j) subseteq cl bigg( cup_m=1^n A_m bigg)$?
This is intuitive, but I was thinking of whether one can be sure that there are no counterexamples.
elementary-set-theory
elementary-set-theory
edited Mar 23 at 18:04
Henning Makholm
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
edited Mar 23 at 18:04
Henning Makholm
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
edited Mar 23 at 18:04
Henning Makholm
243k17310554
edited Mar 23 at 18:04
Henning Makholm
243k17310554
edited Mar 23 at 18:04
Henning Makholm
243k17310554
243k17310554
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
asked Mar 23 at 17:47
mavaviljmavavilj
2,85711138
2,85711138
$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06
add a comment |
$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06
$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06
add a comment |
1 Answer
1
active
oldest
votes
4
$begingroup$
It's clear that $operatornamecl (bigcup_n=1^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
|
show 2 more comments
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$begingroup$
It's clear that $operatornamecl (bigcup_n=1^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
|
show 2 more comments
4
$begingroup$
It's clear that $operatornamecl (bigcup_n=1^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
|
show 2 more comments
4
4
4
$begingroup$
It's clear that $operatornamecl (bigcup_n=1^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
It's clear that $operatornamecl (bigcup_n=1^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
edited Mar 23 at 18:10
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 23 at 18:06
Henno BrandsmaHenno Brandsma
116k349127
116k349127
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
|
show 2 more comments
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
$begingroup$
Why does the union being closed imply that the individual closed subset would be there?
$endgroup$
– mavavilj
Mar 23 at 18:07
1
1
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
@mavavilj: This is completely independent of what is inside the $operatornamecl(cdots)$ on the right-hand-side. The point is simply that the entire right-hand side is one of the things that $operatornamecl(A_j)$ is by definition the smallest of.
$endgroup$
– Henning Makholm
Mar 23 at 18:08
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
$begingroup$
When one takes $cl(A_j)$ then surely one must add the boundary to $A_j$? Then how does one know that the boundary is in the union?
$endgroup$
– mavavilj
Mar 23 at 18:09
1
1
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
@mavavilj: I think you ought to read Henno Brandsma's excellent answer -- instead of just commenting on it without reading what he writes! There is nothing at all about boundaries or unions in this argument.
$endgroup$
– Henning Makholm
Mar 23 at 18:10
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
$begingroup$
If $A_j$ is open, then $cl(A_j) not= A_j$? Or?
$endgroup$
– mavavilj
Mar 23 at 18:11
|
show 2 more comments
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$begingroup$
how do you define closure?
$endgroup$
– Henno Brandsma
Mar 23 at 18:03
$begingroup$
@HennoBrandsma Closure of $K$ is the smallest closed set that contains $K$.
$endgroup$
– mavavilj
Mar 23 at 18:03
$begingroup$
My problem is with that how can I be sure that $cl(A_j)$ would not have points that could lie outside of the union on the R.H.S. Even if points of $A_j$ are in the union.
$endgroup$
– mavavilj
Mar 23 at 18:04
$begingroup$
The closure operation is idempotent and (weakly) order preservering so $Xsubseteq operatornamecl(Y)$ implies $operatornamecl(X)subseteqoperatornamecl(Y)$ in general.
$endgroup$
– Henning Makholm
Mar 23 at 18:06