Let $X = $ $S : S$ is a set and $S notin S$. Is $1, 2$ $in X$?

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

And a follow-up question:
Is $X in X$, where $X$ is as defined previously?

I'm not sure about my reasoning. I think that since no set can be an element of itself, $X$ can contain any set, hence $1, 2$ $in X$.

As for the next question... Well, is this exercise supposed to make one think about Cantor's paradox or am I missing something and there's a "straightforward" answer?

elementary-set-theory

share|cite|improve this question

asked Jul 27 at 20:38

Dumb Dumb

293

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
  – The Count
  Jul 27 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
  – JMoravitz
  Jul 27 at 20:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
  – m_t_
  Jul 27 at 20:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
  – Dog_69
  Jul 27 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you define $1$ and $2$ would have immense repercussions to the answer.
  – Asaf Karagila
  Jul 27 at 21:56
  

  
  
  
  

 | 
show 7 more comments

up vote
0
down vote

favorite

And a follow-up question:
Is $X in X$, where $X$ is as defined previously?

I'm not sure about my reasoning. I think that since no set can be an element of itself, $X$ can contain any set, hence $1, 2$ $in X$.

As for the next question... Well, is this exercise supposed to make one think about Cantor's paradox or am I missing something and there's a "straightforward" answer?

elementary-set-theory

share|cite|improve this question

asked Jul 27 at 20:38

Dumb Dumb

293

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
  – The Count
  Jul 27 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
  – JMoravitz
  Jul 27 at 20:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
  – m_t_
  Jul 27 at 20:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
  – Dog_69
  Jul 27 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you define $1$ and $2$ would have immense repercussions to the answer.
  – Asaf Karagila
  Jul 27 at 21:56
  

  
  
  
  

 | 
show 7 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

And a follow-up question:
Is $X in X$, where $X$ is as defined previously?

I'm not sure about my reasoning. I think that since no set can be an element of itself, $X$ can contain any set, hence $1, 2$ $in X$.

As for the next question... Well, is this exercise supposed to make one think about Cantor's paradox or am I missing something and there's a "straightforward" answer?

elementary-set-theory

share|cite|improve this question

asked Jul 27 at 20:38

Dumb Dumb

293

And a follow-up question:
Is $X in X$, where $X$ is as defined previously?

I'm not sure about my reasoning. I think that since no set can be an element of itself, $X$ can contain any set, hence $1, 2$ $in X$.

As for the next question... Well, is this exercise supposed to make one think about Cantor's paradox or am I missing something and there's a "straightforward" answer?

elementary-set-theory

share|cite|improve this question

asked Jul 27 at 20:38

Dumb Dumb

293

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 20:38

Dumb Dumb

293

asked Jul 27 at 20:38

Dumb Dumb

293

asked Jul 27 at 20:38

Dumb Dumb

293

293

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
  – The Count
  Jul 27 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
  – JMoravitz
  Jul 27 at 20:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
  – m_t_
  Jul 27 at 20:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
  – Dog_69
  Jul 27 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you define $1$ and $2$ would have immense repercussions to the answer.
  – Asaf Karagila
  Jul 27 at 21:56
  

  
  
  
  

 | 
show 7 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
  – The Count
  Jul 27 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
  – JMoravitz
  Jul 27 at 20:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
  – m_t_
  Jul 27 at 20:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
  – Dog_69
  Jul 27 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you define $1$ and $2$ would have immense repercussions to the answer.
  – Asaf Karagila
  Jul 27 at 21:56
  

  
  
  
  

The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
– The Count
Jul 27 at 20:42

The only elements in $1,2$ are $1$ and $2$. It is not a set of sets, so no set can be in it. The second sentence in your second line makes absolutely no sense.
– The Count
Jul 27 at 20:42

2

2

Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
– JMoravitz
Jul 27 at 20:43

Russel's paradox, not Cantor's paradox. Yes, $1,2in X$ (if you allow this notation despite the fact that $X$ is seen not to be a set at all). As for whether $Xin X$ or $Xnotin X$, neither are possible, hence the paradox. The resolution of this paradox is to agree then that $X$ must not be a "set" in the first place, but rather something more exotic.
– JMoravitz
Jul 27 at 20:43

1

1

You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
– m_t_
Jul 27 at 20:44

You are probably intended to check that $1,2 notin X$ directly using the definition, rather than by using the fact that no set is an element of itself. Yes, the second is supposed to make you think about paradoxes.
– m_t_
Jul 27 at 20:44

1

1

@TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
– Dog_69
Jul 27 at 21:48

@TheCount $1$ is defined in set theory as $emptyset$ while $2=emptyset, 1$.Then $1,2$ is actually a set of sets.
– Dog_69
Jul 27 at 21:48

1

1

How do you define $1$ and $2$ would have immense repercussions to the answer.
– Asaf Karagila
Jul 27 at 21:56

How do you define $1$ and $2$ would have immense repercussions to the answer.
– Asaf Karagila
Jul 27 at 21:56

 | 
show 7 more comments

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

To check whether $1,2in X$ you should ask yourself: $$mboxis 1,2in1,2?$$

The second question is a bit more interesting, if $X$ is a set then let's see what are the options;

If $Xin X$ then, by the construction of $X$, we have that $Xnotin X$.

If $Xnotin X$ then $X$ answer the criteria to "enter" $X$, hence $Xin X$.

Both cases leads to construction thus $X$ is not a set but a proper class!

This is indeed suppose to make you think about Russell's paradox(there is a thing called Cantor's paradox but it is different thing, you'll probably learn about it when you'll get to cardinal numbers)

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864744%2flet-x-s-s-is-a-set-and-s-notin-s-is-1-2-in-x%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

To check whether $1,2in X$ you should ask yourself: $$mboxis 1,2in1,2?$$

The second question is a bit more interesting, if $X$ is a set then let's see what are the options;

If $Xin X$ then, by the construction of $X$, we have that $Xnotin X$.

If $Xnotin X$ then $X$ answer the criteria to "enter" $X$, hence $Xin X$.

Both cases leads to construction thus $X$ is not a set but a proper class!

This is indeed suppose to make you think about Russell's paradox(there is a thing called Cantor's paradox but it is different thing, you'll probably learn about it when you'll get to cardinal numbers)

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

add a comment | 

up vote
2
down vote



accepted

To check whether $1,2in X$ you should ask yourself: $$mboxis 1,2in1,2?$$

The second question is a bit more interesting, if $X$ is a set then let's see what are the options;

If $Xin X$ then, by the construction of $X$, we have that $Xnotin X$.

If $Xnotin X$ then $X$ answer the criteria to "enter" $X$, hence $Xin X$.

Both cases leads to construction thus $X$ is not a set but a proper class!

This is indeed suppose to make you think about Russell's paradox(there is a thing called Cantor's paradox but it is different thing, you'll probably learn about it when you'll get to cardinal numbers)

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

To check whether $1,2in X$ you should ask yourself: $$mboxis 1,2in1,2?$$

The second question is a bit more interesting, if $X$ is a set then let's see what are the options;

If $Xin X$ then, by the construction of $X$, we have that $Xnotin X$.

If $Xnotin X$ then $X$ answer the criteria to "enter" $X$, hence $Xin X$.

Both cases leads to construction thus $X$ is not a set but a proper class!

This is indeed suppose to make you think about Russell's paradox(there is a thing called Cantor's paradox but it is different thing, you'll probably learn about it when you'll get to cardinal numbers)

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

To check whether $1,2in X$ you should ask yourself: $$mboxis 1,2in1,2?$$

The second question is a bit more interesting, if $X$ is a set then let's see what are the options;

If $Xin X$ then, by the construction of $X$, we have that $Xnotin X$.

If $Xnotin X$ then $X$ answer the criteria to "enter" $X$, hence $Xin X$.

Both cases leads to construction thus $X$ is not a set but a proper class!

This is indeed suppose to make you think about Russell's paradox(there is a thing called Cantor's paradox but it is different thing, you'll probably learn about it when you'll get to cardinal numbers)

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 21:43

Holo

4,1072528

answered Jul 27 at 21:43

Holo

4,1072528

answered Jul 27 at 21:43

Holo

4,1072528

4,1072528

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864744%2flet-x-s-s-is-a-set-and-s-notin-s-is-1-2-in-x%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email