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7

$begingroup$

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

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edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

$endgroup$

add a comment | 

7

$begingroup$

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

share|cite|improve this question

edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

$endgroup$

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

share|cite|improve this question

edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

$endgroup$

share|cite|improve this question

edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

share|cite|improve this question

edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

edited 1 hour ago

LarsH

555624

edited 1 hour ago

LarsH

555624

asked 8 hours ago

Ica SanduIca Sandu

1379

asked 8 hours ago

Ica SanduIca Sandu

1379

4 Answers
4

active

oldest

votes

22

$begingroup$

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 $emptysetin{1,emptyset}$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

CornmanCornman

3,69321231

$endgroup$

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 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

add a comment | 

5

$begingroup$

The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

CiaPanCiaPan

10.3k11248

$endgroup$

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 8 hours ago

MelodyMelody

1,21312

$endgroup$

add a comment | 

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22

$begingroup$

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 $emptysetin{1,emptyset}$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

CornmanCornman

3,69321231

$endgroup$

add a comment | 

22

$begingroup$

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 $emptysetin{1,emptyset}$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

CornmanCornman

3,69321231

$endgroup$

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 $emptysetin{1,emptyset}$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

CornmanCornman

3,69321231

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

CornmanCornman

3,69321231

answered 8 hours ago

CornmanCornman

3,69321231

answered 8 hours ago

CornmanCornman

3,69321231

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 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

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 8 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).

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

5

$begingroup$

The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

CiaPanCiaPan

10.3k11248

$endgroup$

add a comment | 

5

$begingroup$

The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

CiaPanCiaPan

10.3k11248

$endgroup$

add a comment | 

$begingroup$

The answer is: the given definition uses $cup{emptyset} $, not $cupemptyset $, so it adds the empty set as an element, not a subset of $S $.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

CiaPanCiaPan

10.3k11248

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

CiaPanCiaPan

10.3k11248

answered 6 hours ago

CiaPanCiaPan

10.3k11248

answered 6 hours ago

CiaPanCiaPan

10.3k11248

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 8 hours ago

MelodyMelody

1,21312

$endgroup$

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 8 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 $ainemptyset.$

share|cite|improve this answer

answered 8 hours ago

MelodyMelody

1,21312

$endgroup$

share|cite|improve this answer

answered 8 hours ago

MelodyMelody

1,21312

answered 8 hours ago

MelodyMelody

1,21312

answered 8 hours ago

MelodyMelody

1,21312