Gsyfuy

Wall plug outlet change

Finding the path in a graph from A to B then back to A with a minimum of shared edges

Did the new image of black hole confirm the general theory of relativity?

Change bounding box of math glyphs in LuaTeX

What was the last x86 CPU that did not have the x87 floating-point unit built in?

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

Working through the single responsibility principle (SRP) in Python when calls are expensive

The variadic template constructor of my class cannot modify my class members, why is that so?

Was credit for the black hole image misattributed?

What's the point in a preamp?

How are presidential pardons supposed to be used?

Create an outline of font

Can withdrawing asylum be illegal?

Hopping to infinity along a string of digits

Sort list of array linked objects by keys and values

Is this wall load bearing? Blueprints and photos attached

How to grep and cut numbers from a file and sum them

How is simplicity better than precision and clarity in prose?

Are spiders unable to hurt humans, especially very small spiders?

Segmentation fault output is suppressed when piping stdin into a function. Why?

Windows 10: How to Lock (not sleep) laptop on lid close?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

I could not break this equation. Please help me

Does Parliament hold absolute power in the UK?

If the empty set is a subset of every set, why write … ∪ {∅}?

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 32 mins ago

LarsH

555624

asked 7 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 32 mins ago

LarsH

555624

asked 7 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 32 mins ago

LarsH

555624

asked 7 hours ago

Ica SanduIca Sandu

1379

$endgroup$

share|cite|improve this question

edited 32 mins ago

LarsH

555624

asked 7 hours ago

Ica SanduIca Sandu

1379

share|cite|improve this question

edited 32 mins ago

LarsH

555624

asked 7 hours ago

Ica SanduIca Sandu

1379

edited 32 mins ago

LarsH

555624

edited 32 mins ago

LarsH

555624

asked 7 hours ago

Ica SanduIca Sandu

1379

asked 7 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 5 hours ago

answered 7 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 7 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 5 hours ago

answered 5 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 7 hours ago

MelodyMelody

1,21312

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3186480%2fif-the-empty-set-is-a-subset-of-every-set-why-write-%25e2%2588%25aa-%25e2%2588%2585%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 5 hours ago

answered 7 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 5 hours ago

answered 7 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 5 hours ago

answered 7 hours ago

CornmanCornman

3,69321231

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 7 hours ago

CornmanCornman

3,69321231

answered 7 hours ago

CornmanCornman

3,69321231

answered 7 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 7 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 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).

share|cite|improve this answer

answered 7 hours ago

José Carlos SantosJosé Carlos Santos

174k23134243

$endgroup$

share|cite|improve this answer

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

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 5 hours ago

answered 5 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 5 hours ago

answered 5 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 5 hours ago

answered 5 hours ago

CiaPanCiaPan

10.3k11248

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

CiaPanCiaPan

10.3k11248

answered 5 hours ago

CiaPanCiaPan

10.3k11248

answered 5 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 7 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 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 $ainemptyset.$

share|cite|improve this answer

answered 7 hours ago

MelodyMelody

1,21312

$endgroup$

share|cite|improve this answer

answered 7 hours ago

MelodyMelody

1,21312

answered 7 hours ago

MelodyMelody

1,21312

answered 7 hours ago

MelodyMelody

1,21312