If the empty set is a subset of every set, why write … $cup {∅}$?
$begingroup$
I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $
I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?
measure-theory elementary-set-theory notation
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
$endgroup$
add a comment |
$begingroup$
I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $
I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?
measure-theory elementary-set-theory notation
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
$endgroup$
$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday
add a comment |
$begingroup$
I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $
I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?
measure-theory elementary-set-theory notation
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
$endgroup$
I met the notation $ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $
I know $S$ is a family of subsets ,a set of intervals, and from set theory $emptyset$ is a subsets of every set then why in the notation :$ S={(a,b] ; a,bin mathbb R,a<b}cup{emptyset} $ appear $color{red}{cup{emptyset}}$?
measure-theory elementary-set-theory notation
measure-theory elementary-set-theory notation
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
edited Apr 14 at 1:59
YuiTo Cheng
2,54841037
2,54841037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
1399
$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday
add a comment |
$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday
$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday
$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
answered Apr 13 at 17:03
CornmanCornman
3,81021233
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, so it adds the empty set as an element, not a subset of $S $.
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
add a comment |
$begingroup$
It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.
As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$
answered Apr 13 at 17:05
MelodyMelody
1,31212
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
answered Apr 13 at 17:03
CornmanCornman
3,81021233
$endgroup$
add a comment |
$begingroup$
It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
answered Apr 13 at 17:03
CornmanCornman
3,81021233
$endgroup$
add a comment |
$begingroup$
It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
answered Apr 13 at 17:03
CornmanCornman
3,81021233
$endgroup$
It is because the emptyset $emptyset$ is a subset of every set, but not an element of every set.
It is $emptysetin S$ and you might want that to show, that the elements of $S$ define a topology.
Or to be more clear it is ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
answered Apr 13 at 17:03
CornmanCornman
3,81021233
edited Apr 13 at 18:36
answered Apr 13 at 17:03
CornmanCornman
3,81021233
answered Apr 13 at 17:03
CornmanCornman
3,81021233
answered Apr 13 at 17:03
CornmanCornman
3,81021233
3,81021233
add a comment |
add a comment |
$begingroup$
The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, so it adds the empty set as an element, not a subset of $S $.
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, so it adds the empty set as an element, not a subset of $S $.
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, so it adds the empty set as an element, not a subset of $S $.
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
$endgroup$
The answer is: the given definition uses $cup,{emptyset} $, not $cup,emptyset $, so it adds the empty set as an element, not a subset of $S $.
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
edited Apr 15 at 10:39
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
answered Apr 13 at 18:57
CiaPanCiaPan
10.4k11248
10.4k11248
add a comment |
add a comment |
$begingroup$
Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
add a comment |
$begingroup$
Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
add a comment |
$begingroup$
Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
Because the empty set $(emptyset)$ is one thing, but what you have there is ${emptyset}$, which is a different thing: it's a set with a single element (which happens to be the empty set).
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
176k24135244
176k24135244
add a comment |
add a comment |
$begingroup$
It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.
As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$
answered Apr 13 at 17:05
MelodyMelody
1,31212
$endgroup$
add a comment |
$begingroup$
It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.
As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$
answered Apr 13 at 17:05
MelodyMelody
1,31212
$endgroup$
add a comment |
$begingroup$
It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.
As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$
answered Apr 13 at 17:05
MelodyMelody
1,31212
$endgroup$
It looks like $S$ is denoting subintervals of the real line that are open on the left and closed on the right with the convention that $emptyset$ is such a subinterval. In which case there is nothing to show, it's just a convention that $emptyset$ is a subinterval. The reason for using ${emptyset}$ is show you can write out the collection of all such subintervals in a nice form.
As for the empty set is a subset of every set, well that's a vacuous truth. For all $ainemptyset$ if $X$ is a set it follows that $ain X.$ This is true, because there are no $ainemptyset.$
answered Apr 13 at 17:05
MelodyMelody
1,31212
answered Apr 13 at 17:05
MelodyMelody
1,31212
answered Apr 13 at 17:05
MelodyMelody
1,31212
answered Apr 13 at 17:05
MelodyMelody
1,31212
1,31212
add a comment |
add a comment |
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$begingroup$
A way of understanding this is to write $varnothing$ as ${}$ and ${varnothing}$ as ${{}}$.
$endgroup$
– Trebor
yesterday