If the empty set is a subset of every set, why write … ∪ {∅}? The 2019 Stack Overflow...
How is simplicity better than precision and clarity in prose?
Is there a writing software that you can sort scenes like slides in PowerPoint?
Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?
Are spiders unable to hurt humans, especially very small spiders?
I could not break this equation. Please help me
What's the point in a preamp?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Road tyres vs "Street" tyres for charity ride on MTB Tandem
What is special about square numbers here?
The variadic template constructor of my class cannot modify my class members, why is that so?
Relations between two reciprocal partial derivatives?
How did passengers keep warm on sail ships?
How does this infinite series simplify to an integral?
Can withdrawing asylum be illegal?
A pet rabbit called Belle
Can a novice safely splice in wire to lengthen 5V charging cable?
Semisimplicity of the category of coherent sheaves?
Arduino Pro Micro - switch off LEDs
Python - Fishing Simulator
Is this wall load bearing? Blueprints and photos attached
How to test the equality of two Pearson correlation coefficients computed from the same sample?
When did F become S in typeography, and why?
Keeping a retro style to sci-fi spaceships?
Scientific Reports - Significant Figures
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 {$xin A | p(x)$}?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<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
$endgroup$
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
$endgroup$
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
$endgroup$
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
measure-theory elementary-set-theory
edited 14 mins ago
LarsH
555624
555624
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 ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
$endgroup$
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).
$endgroup$
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 $.
$endgroup$
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 $ainemptyset.$
$endgroup$
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
});
}
});
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
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
Required, but never shown
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 ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
$endgroup$
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 $emptysetin{1,emptyset}$
$endgroup$
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 $emptysetin{1,emptyset}$
$endgroup$
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 ${1}neq{1,emptyset}$. The set on the left has one element, the set on the right has two elements, with $emptysetin{1,emptyset}$
edited 5 hours ago
answered 7 hours ago
CornmanCornman
3,69321231
3,69321231
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).
$endgroup$
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).
$endgroup$
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).
$endgroup$
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
174k23134243
add a comment |
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 $.
$endgroup$
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 $.
$endgroup$
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 $.
$endgroup$
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 $.
edited 4 hours ago
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 $ainemptyset.$
$endgroup$
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 $ainemptyset.$
$endgroup$
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 $ainemptyset.$
$endgroup$
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
1,21312
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown