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If the empty set is a subset of every set, why write … $cup ∅$?
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 "to every set and to every statement p(x), there exists $xin A ?Should the empty set be included in this example?What subset am I missing from a set containing the empty set and a set with the empty set?Union on the empty set and the set containing the empty setWhy the empty set is a subset of every set?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 notation
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
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<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 notation
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
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
19 hours ago
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 notation
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
$endgroup$
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 notation
measure-theory elementary-set-theory notation
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
asked Apr 13 at 17:00
Ica SanduIca Sandu
1399
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
edited Apr 14 at 1:59
YuiTo Cheng
2,52341037
2,52341037
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
19 hours ago
add a comment |
$begingroup$
A way of understanding this is to write $varnothing$ as $$ and $varnothing$ as $$.
$endgroup$
– Trebor
19 hours ago
$begingroup$
A way of understanding this is to write $varnothing$ as $$ and $varnothing$ as $$.
$endgroup$
– Trebor
19 hours ago
$begingroup$
A way of understanding this is to write $varnothing$ as $$ and $varnothing$ as $$.
$endgroup$
– Trebor
19 hours ago
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 Apr 13 at 17:03
CornmanCornman
3,77021233
$endgroup$
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
175k24134243
$endgroup$
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
1,30712
$endgroup$
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 Apr 13 at 17:03
CornmanCornman
3,77021233
$endgroup$
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 Apr 13 at 17:03
CornmanCornman
3,77021233
$endgroup$
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 Apr 13 at 17:03
CornmanCornman
3,77021233
$endgroup$
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 Apr 13 at 17:03
CornmanCornman
3,77021233
edited Apr 13 at 18:36
answered Apr 13 at 17:03
CornmanCornman
3,77021233
answered Apr 13 at 17:03
CornmanCornman
3,77021233
answered Apr 13 at 17:03
CornmanCornman
3,77021233
3,77021233
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 2 days ago
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
175k24134243
$endgroup$
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
175k24134243
$endgroup$
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
175k24134243
$endgroup$
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
175k24134243
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 13 at 17:03
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
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
1,30712
$endgroup$
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
1,30712
$endgroup$
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
1,30712
$endgroup$
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
1,30712
answered Apr 13 at 17:05
MelodyMelody
1,30712
answered Apr 13 at 17:05
MelodyMelody
1,30712
answered Apr 13 at 17:05
MelodyMelody
1,30712
1,30712
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
19 hours ago