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If the empty set is a subset of every set, why write … $cup ∅$?

7

1

$begingroup$

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

share|cite|improve this question

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
  12 hours ago
  

  
  
  
  

add a comment | 

7

1

$begingroup$

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

share|cite|improve this question

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
  12 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

share|cite|improve this question

edited Apr 14 at 1:59

YuiTo Cheng

2,52341037

asked Apr 13 at 17:00

Ica SanduIca Sandu

1399

$endgroup$

share|cite|improve this question

edited Apr 14 at 1:59

YuiTo Cheng

2,52341037

asked Apr 13 at 17:00

Ica SanduIca Sandu

1399

share|cite|improve this question

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

asked Apr 13 at 17:00

Ica SanduIca Sandu

1399

asked Apr 13 at 17:00

Ica SanduIca Sandu

1399

4 Answers
4

active

oldest

votes

24

$begingroup$

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 $emptysetin1,emptyset$

share|cite|improve this answer

edited Apr 13 at 18:36

answered Apr 13 at 17:03

CornmanCornman

3,77021233

$endgroup$

add a comment | 

6

$begingroup$

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 $.

share|cite|improve this answer

edited 2 days ago

answered Apr 13 at 18:57

CiaPanCiaPan

10.4k11248

$endgroup$

add a comment | 

5

$begingroup$

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).

share|cite|improve this answer

answered Apr 13 at 17:03

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

add a comment | 

3

$begingroup$

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 $ainemptyset.$

share|cite|improve this answer

answered Apr 13 at 17:05

MelodyMelody

1,30012

$endgroup$

add a comment | 

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24

$begingroup$

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 $emptysetin1,emptyset$

share|cite|improve this answer

edited Apr 13 at 18:36

answered Apr 13 at 17:03

CornmanCornman

3,77021233

$endgroup$

add a comment | 

24

$begingroup$

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 $emptysetin1,emptyset$

share|cite|improve this answer

edited Apr 13 at 18:36

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 $emptysetin1,emptyset$

share|cite|improve this answer

edited Apr 13 at 18:36

answered Apr 13 at 17:03

CornmanCornman

3,77021233

$endgroup$

share|cite|improve this answer

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

6

$begingroup$

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 $.

share|cite|improve this answer

edited 2 days ago

answered Apr 13 at 18:57

CiaPanCiaPan

10.4k11248

$endgroup$

add a comment | 

6

$begingroup$

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 $.

share|cite|improve this answer

edited 2 days ago

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 $.

share|cite|improve this answer

edited 2 days ago

answered Apr 13 at 18:57

CiaPanCiaPan

10.4k11248

$endgroup$

share|cite|improve this answer

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

5

$begingroup$

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).

share|cite|improve this answer

answered Apr 13 at 17:03

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

add a comment | 

5

$begingroup$

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).

share|cite|improve this answer

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).

share|cite|improve this answer

answered Apr 13 at 17:03

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

share|cite|improve this answer

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

3

$begingroup$

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 $ainemptyset.$

share|cite|improve this answer

answered Apr 13 at 17:05

MelodyMelody

1,30012

$endgroup$

add a comment | 

3

$begingroup$

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 $ainemptyset.$

share|cite|improve this answer

answered Apr 13 at 17:05

MelodyMelody

1,30012

$endgroup$

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 $ainemptyset.$

share|cite|improve this answer

answered Apr 13 at 17:05

MelodyMelody

1,30012

$endgroup$

share|cite|improve this answer

answered Apr 13 at 17:05

MelodyMelody

1,30012

answered Apr 13 at 17:05

MelodyMelody

1,30012

answered Apr 13 at 17:05

MelodyMelody

1,30012