If the empty set is a subset of every set, why write … ∪ ∅? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the void set (∅) a proper subset of every set?Direct proof of empty set being subset of every setIf the empty set is a subset of every set, why isn't $emptyset,a=a$?Why $ emptyset $ is not a subset of $ emptyset $A set $X$ is called 'complete' if every element of $X$ is subset of $X$.Why "to every set and to every statement p(x), there exists $xin A ?What subset am I missing from a set containing the empty set and a set with the empty set?Does an element of a set, that can't be in a list, make that set uncountable?Question about the empty setUnderstanding empty set, set with empty set and set with set of empty set.
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If the empty set is a subset of every set, why write … ∪ ∅?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the void set (∅) a proper subset of every set?Direct proof of empty set being subset of every setIf the empty set is a subset of every set, why isn't $emptyset,a=a$?Why $ emptyset $ is not a subset of $ emptyset $A set $X$ is called 'complete' if every element of $X$ is subset of $X$.Why "to every set and to every statement p(x), there exists p(x)$?What subset am I missing from a set containing the empty set and a set with the empty set?Does an element of a set, that can't be in a list, make that set uncountable?Question about the empty setUnderstanding empty set, set with empty set and set with set of empty set.
$begingroup$
I met the notation $ S=(a,b] ; a,bin mathbb R,a<bcupemptyset $
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<bcupemptyset $ appear $colorredcupemptyset$?
measure-theory elementary-set-theory
edited 8 mins ago
LarsH
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
$endgroup$
add a comment |
$begingroup$
I met the notation $ S=(a,b] ; a,bin mathbb R,a<bcupemptyset $
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<bcupemptyset $ appear $colorredcupemptyset$?
measure-theory elementary-set-theory
edited 8 mins ago
LarsH
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
$endgroup$
add a comment |
$begingroup$
I met the notation $ S=(a,b] ; a,bin mathbb R,a<bcupemptyset $
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<bcupemptyset $ appear $colorredcupemptyset$?
measure-theory elementary-set-theory
edited 8 mins ago
LarsH
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
$endgroup$
I met the notation $ S=(a,b] ; a,bin mathbb R,a<bcupemptyset $
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<bcupemptyset $ appear $colorredcupemptyset$?
measure-theory elementary-set-theory
measure-theory elementary-set-theory
edited 8 mins ago
LarsH
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
edited 8 mins ago
LarsH
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
edited 8 mins ago
LarsH
555624
edited 8 mins ago
LarsH
555624
edited 8 mins ago
LarsH
555624
555624
asked 7 hours ago
Ica SanduIca Sandu
1329
asked 7 hours ago
Ica SanduIca Sandu
1329
asked 7 hours ago
Ica SanduIca Sandu
1329
1329
add a comment |
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 $1neq1,emptyset$. The set on the left has one element, the set on the right has two elements, with $emptysetin1,emptyset$
answered 7 hours ago
CornmanCornman
3,68321231
$endgroup$
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 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cupemptyset $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.
answered 5 hours ago
CiaPanCiaPan
10.3k11248
$endgroup$
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 7 hours ago
MelodyMelody
1,21312
$endgroup$
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 $1neq1,emptyset$. The set on the left has one element, the set on the right has two elements, with $emptysetin1,emptyset$
answered 7 hours ago
CornmanCornman
3,68321231
$endgroup$
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 $1neq1,emptyset$. The set on the left has one element, the set on the right has two elements, with $emptysetin1,emptyset$
answered 7 hours ago
CornmanCornman
3,68321231
$endgroup$
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 $1neq1,emptyset$. The set on the left has one element, the set on the right has two elements, with $emptysetin1,emptyset$
answered 7 hours ago
CornmanCornman
3,68321231
$endgroup$
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 $1neq1,emptyset$. The set on the left has one element, the set on the right has two elements, with $emptysetin1,emptyset$
answered 7 hours ago
CornmanCornman
3,68321231
edited 5 hours ago
answered 7 hours ago
CornmanCornman
3,68321231
answered 7 hours ago
CornmanCornman
3,68321231
answered 7 hours ago
CornmanCornman
3,68321231
3,68321231
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 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
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 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
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 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
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 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
answered 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
answered 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
answered 7 hours ago
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
add a comment |
add a comment |
$begingroup$
The answer is: the given definition uses $cupemptyset $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.
answered 5 hours ago
CiaPanCiaPan
10.3k11248
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cupemptyset $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.
answered 5 hours ago
CiaPanCiaPan
10.3k11248
$endgroup$
add a comment |
$begingroup$
The answer is: the given definition uses $cupemptyset $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.
answered 5 hours ago
CiaPanCiaPan
10.3k11248
$endgroup$
The answer is: the given definition uses $cupemptyset $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.
answered 5 hours ago
CiaPanCiaPan
10.3k11248
edited 4 hours ago
answered 5 hours ago
CiaPanCiaPan
10.3k11248
answered 5 hours ago
CiaPanCiaPan
10.3k11248
answered 5 hours ago
CiaPanCiaPan
10.3k11248
10.3k11248
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 7 hours ago
MelodyMelody
1,21312
$endgroup$
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 7 hours ago
MelodyMelody
1,21312
$endgroup$
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 7 hours ago
MelodyMelody
1,21312
$endgroup$
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 7 hours ago
MelodyMelody
1,21312
answered 7 hours ago
MelodyMelody
1,21312
answered 7 hours ago
MelodyMelody
1,21312
answered 7 hours ago
MelodyMelody
1,21312
1,21312
add a comment |
add a comment |
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