Fdtxjr

Crossing the line between justified force and brutality

Customer Requests (Sometimes) Drive Me Bonkers!

What is the best translation for "slot" in the context of multiplayer video games?

What is the intuitive meaning of having a linear relationship between the logs of two variables?

Type int? vs type int

How to run a prison with the smallest amount of guards?

Go Pregnant or Go Home

How to Reset Passwords on Multiple Websites Easily?

Do sorcerers' Subtle Spells require a skill check to be unseen?

Why escape if the_content isnt?

How to draw lines on a tikz-cd diagram

Lay out the Carpet

India just shot down a satellite from the ground. At what altitude range is the resulting debris field?

How do I find the solutions of the following equation?

How did Doctor Strange see the winning outcome in Avengers: Infinity War?

Large drywall patch supports

What is the opposite of 'gravitas'?

Anatomically Correct Strange Women In Ponds Distributing Swords

Is HostGator storing my password in plaintext?

Detecting if an element is found inside a container

Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?

How long to clear the 'suck zone' of a turbofan after start is initiated?

How does buying out courses with grant money work?

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

If $X$ is a set of ordinals then so is $bigcup X$

Ordinals with the same cardinalityRelation between two distinct class of ordinals$bigcup X$ = $sup(X)$ for a set $X$ of ordinalsElementary Set Theory: Ordinals, Well Orderings and Isomorphismsconfusion about the definition of cardinalityIf $X$ is well-ordered then $Xcup left infty right$ is well-ordered.If $x$ is a set of ordinals, then $bigcup x=alpha$. Why is $xsubseteq alpha+1$ rather than $alpha$Class of Ordinals $On$ not a SetSuppose that $X$ is a set of ordinals. Then there exists an ordinal $betanotin X$Fundemental Properties of Ordinals

1

$begingroup$

Theorem. If $X$ is a set of ordinals then so is $bigcup X.$

Claim 1. $bigcup X$ is well-ordered.

Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.

Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$

Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$

Proof of theorem. By the Claim 1 and the Claim 2.

May you check my proof? Thanks...

elementary-set-theory proof-writing

share|cite|improve this question

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
  $endgroup$
  – Berci
  Mar 17 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Berci I couldn't understand your notation. Can you explain in detail?
  $endgroup$
  – PozcuKushimotoStreet
  Mar 17 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
  $endgroup$
  – Berci
  Mar 18 at 12:02
  

  
  
  
  

add a comment | 

1

$begingroup$

Theorem. If $X$ is a set of ordinals then so is $bigcup X.$

Claim 1. $bigcup X$ is well-ordered.

Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.

Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$

Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$

Proof of theorem. By the Claim 1 and the Claim 2.

May you check my proof? Thanks...

elementary-set-theory proof-writing

share|cite|improve this question

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
  $endgroup$
  – Berci
  Mar 17 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Berci I couldn't understand your notation. Can you explain in detail?
  $endgroup$
  – PozcuKushimotoStreet
  Mar 17 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
  $endgroup$
  – Berci
  Mar 18 at 12:02
  

  
  
  
  

add a comment | 

$begingroup$

Theorem. If $X$ is a set of ordinals then so is $bigcup X.$

Claim 1. $bigcup X$ is well-ordered.

Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.

Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$

Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$

Proof of theorem. By the Claim 1 and the Claim 2.

May you check my proof? Thanks...

elementary-set-theory proof-writing

share|cite|improve this question

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

$endgroup$

share|cite|improve this question

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

share|cite|improve this question

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

asked Mar 17 at 21:28

PozcuKushimotoStreetPozcuKushimotoStreet

1,408923

1 Answer
1

active

oldest

votes

1

$begingroup$

So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.

However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.

Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.

The proof of claim 2 is correct.

share|cite|improve this answer

answered Mar 18 at 14:56

BerciBerci

61.8k23674

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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%2f3152109%2fif-x-is-a-set-of-ordinals-then-so-is-bigcup-x%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.

However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.

Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.

The proof of claim 2 is correct.

share|cite|improve this answer

answered Mar 18 at 14:56

BerciBerci

61.8k23674

$endgroup$

add a comment | 

1

$begingroup$

So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.

However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.

Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.

The proof of claim 2 is correct.

share|cite|improve this answer

answered Mar 18 at 14:56

BerciBerci

61.8k23674

$endgroup$

add a comment | 

$begingroup$

So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.

However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.

Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.

The proof of claim 2 is correct.

share|cite|improve this answer

answered Mar 18 at 14:56

BerciBerci

61.8k23674

$endgroup$

share|cite|improve this answer

answered Mar 18 at 14:56

BerciBerci

61.8k23674

answered Mar 18 at 14:56

BerciBerci

61.8k23674

answered Mar 18 at 14:56

BerciBerci

61.8k23674