Does $liminf$ distribute?
$begingroup$
I know the property that for sets $A_n,B_n$
$$
limsup_{ntoinfty}(A_ncup B_n)=limsup_{ntoinfty}(A_n)cup limsup_{ntoinfty}(B_n)
$$
holds.
I'm curious if it also holds that
$$
liminf_{ntoinfty}(A_ncup B_n)=liminf_{ntoinfty}(A_n)cup liminf_{ntoinfty}(B_n)?
$$
limits elementary-set-theory limsup-and-liminf
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
$endgroup$
add a comment |
$begingroup$
I know the property that for sets $A_n,B_n$
$$
limsup_{ntoinfty}(A_ncup B_n)=limsup_{ntoinfty}(A_n)cup limsup_{ntoinfty}(B_n)
$$
holds.
I'm curious if it also holds that
$$
liminf_{ntoinfty}(A_ncup B_n)=liminf_{ntoinfty}(A_n)cup liminf_{ntoinfty}(B_n)?
$$
limits elementary-set-theory limsup-and-liminf
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
$endgroup$
$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
1
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48
add a comment |
$begingroup$
I know the property that for sets $A_n,B_n$
$$
limsup_{ntoinfty}(A_ncup B_n)=limsup_{ntoinfty}(A_n)cup limsup_{ntoinfty}(B_n)
$$
holds.
I'm curious if it also holds that
$$
liminf_{ntoinfty}(A_ncup B_n)=liminf_{ntoinfty}(A_n)cup liminf_{ntoinfty}(B_n)?
$$
limits elementary-set-theory limsup-and-liminf
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
$endgroup$
I know the property that for sets $A_n,B_n$
$$
limsup_{ntoinfty}(A_ncup B_n)=limsup_{ntoinfty}(A_n)cup limsup_{ntoinfty}(B_n)
$$
holds.
I'm curious if it also holds that
$$
liminf_{ntoinfty}(A_ncup B_n)=liminf_{ntoinfty}(A_n)cup liminf_{ntoinfty}(B_n)?
$$
limits elementary-set-theory limsup-and-liminf
limits elementary-set-theory limsup-and-liminf
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
asked Dec 10 '18 at 1:31
BrutusBrutus
147110
147110
$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
1
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48
add a comment |
$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
1
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48
$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
1
1
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$limsup_{n to infty} (A_n cup B_n)$ contains elements in $A_n cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $liminf$ suggests you need to consider intersections instead of unions.
$liminf_{n to infty} (A_n cap B_n)$ contains elements in $A_n cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$liminf_{n to infty} (A_n cap B_n)
= liminf_{n to infty} (A_n) cap liminf_{n to infty} (B_n).$$
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
$endgroup$
add a comment |
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$begingroup$
$limsup_{n to infty} (A_n cup B_n)$ contains elements in $A_n cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $liminf$ suggests you need to consider intersections instead of unions.
$liminf_{n to infty} (A_n cap B_n)$ contains elements in $A_n cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$liminf_{n to infty} (A_n cap B_n)
= liminf_{n to infty} (A_n) cap liminf_{n to infty} (B_n).$$
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
$endgroup$
add a comment |
$begingroup$
$limsup_{n to infty} (A_n cup B_n)$ contains elements in $A_n cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $liminf$ suggests you need to consider intersections instead of unions.
$liminf_{n to infty} (A_n cap B_n)$ contains elements in $A_n cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$liminf_{n to infty} (A_n cap B_n)
= liminf_{n to infty} (A_n) cap liminf_{n to infty} (B_n).$$
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
$endgroup$
add a comment |
$begingroup$
$limsup_{n to infty} (A_n cup B_n)$ contains elements in $A_n cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $liminf$ suggests you need to consider intersections instead of unions.
$liminf_{n to infty} (A_n cap B_n)$ contains elements in $A_n cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$liminf_{n to infty} (A_n cap B_n)
= liminf_{n to infty} (A_n) cap liminf_{n to infty} (B_n).$$
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
$endgroup$
$limsup_{n to infty} (A_n cup B_n)$ contains elements in $A_n cup B_n$ for infinitely many $n$. These elements are precisely the elements that are either in $A_n$ for infinitely many $n$, or in $B_n$ for infinitely many $n$.
Modifying this argument to $liminf$ suggests you need to consider intersections instead of unions.
$liminf_{n to infty} (A_n cap B_n)$ contains elements in $A_n cap B_n$ for all but finitely many $n$. These elements are precisely the elements that are both in $A_n$ for all but finitely many $n$, and in $B_n$ for all but finitely many $n$.
$$liminf_{n to infty} (A_n cap B_n)
= liminf_{n to infty} (A_n) cap liminf_{n to infty} (B_n).$$
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
edited Dec 10 '18 at 2:10
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
answered Dec 10 '18 at 1:43
angryavianangryavian
41.1k23380
41.1k23380
add a comment |
add a comment |
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$begingroup$
How is the former established?
$endgroup$
– Robert Wolfe
Dec 10 '18 at 1:36
1
$begingroup$
No. Consider $A_n$ which is empty for odd $n$ and equal to ${1}$ for even $n$, and $B_n$ which is empty for even $n$ and equal to ${1}$ for odd $n$. Then $A_ncup B_n={1}$ for all $n$, so the limit inferior is ${1}$; however, each of the limits inferior on the right are empty.
$endgroup$
– Arturo Magidin
Dec 10 '18 at 1:48