Is ∅ ⊈ { ∅, 1, 2 } False?

0

$begingroup$

Is this ∅ ⊈ { ∅, 1, 2 } true or false ?

Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?

Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?

elementary-set-theory

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
  $endgroup$
  – Le Anh Dung
  Jan 21 at 5:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't confuse $in$ with $subseteq$.
  $endgroup$
  – Cheerful Parsnip
  Jan 21 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $emptyset subset A $ is always true no matter what set $A $ is.
  $endgroup$
  – fleablood
  Jan 21 at 7:21
  

  
  
  
  

add a comment | 

0

$begingroup$

Is this ∅ ⊈ { ∅, 1, 2 } true or false ?

Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?

Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?

elementary-set-theory

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
  $endgroup$
  – Le Anh Dung
  Jan 21 at 5:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't confuse $in$ with $subseteq$.
  $endgroup$
  – Cheerful Parsnip
  Jan 21 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $emptyset subset A $ is always true no matter what set $A $ is.
  $endgroup$
  – fleablood
  Jan 21 at 7:21
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Is this ∅ ⊈ { ∅, 1, 2 } true or false ?

Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?

Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?

elementary-set-theory

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

$endgroup$

Is this ∅ ⊈ { ∅, 1, 2 } true or false ?

Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?

Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?

elementary-set-theory

elementary-set-theory

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

share|cite|improve this question

share|cite|improve this question

asked Jan 21 at 5:44

J.SJ.S

555

asked Jan 21 at 5:44

J.SJ.S

555

asked Jan 21 at 5:44

J.SJ.S

555

555

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
  $endgroup$
  – Le Anh Dung
  Jan 21 at 5:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't confuse $in$ with $subseteq$.
  $endgroup$
  – Cheerful Parsnip
  Jan 21 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $emptyset subset A $ is always true no matter what set $A $ is.
  $endgroup$
  – fleablood
  Jan 21 at 7:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
  $endgroup$
  – Le Anh Dung
  Jan 21 at 5:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't confuse $in$ with $subseteq$.
  $endgroup$
  – Cheerful Parsnip
  Jan 21 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $emptyset subset A $ is always true no matter what set $A $ is.
  $endgroup$
  – fleablood
  Jan 21 at 7:21
  

  
  
  
  

$begingroup$
It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
$endgroup$
– Le Anh Dung
Jan 21 at 5:47

$begingroup$
It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
$endgroup$
– Le Anh Dung
Jan 21 at 5:47

2

2

$begingroup$
Don't confuse $in$ with $subseteq$.
$endgroup$
– Cheerful Parsnip
Jan 21 at 5:55

$begingroup$
Don't confuse $in$ with $subseteq$.
$endgroup$
– Cheerful Parsnip
Jan 21 at 5:55

$begingroup$
$emptyset subset A $ is always true no matter what set $A $ is.
$endgroup$
– fleablood
Jan 21 at 7:21

$begingroup$
$emptyset subset A $ is always true no matter what set $A $ is.
$endgroup$
– fleablood
Jan 21 at 7:21

add a comment | 

3 Answers
3

active

oldest

votes

2

$begingroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, why is {⊘} ⊂ {⊘,1,2} ?
  $endgroup$
  – J.S
  Jan 21 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
  $endgroup$
  – Anik Bhowmick
  Jan 21 at 5:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.S Because $oslashin{oslash,1,2}.$
  $endgroup$
  – Michael Rozenberg
  Jan 21 at 5:54
  

  
  
  
  

add a comment | 

0

$begingroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

$endgroup$

add a comment | 

0

$begingroup$

$neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.

Consider the case $A=phi.$

Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, why is {⊘} ⊂ {⊘,1,2} ?
  $endgroup$
  – J.S
  Jan 21 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
  $endgroup$
  – Anik Bhowmick
  Jan 21 at 5:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.S Because $oslashin{oslash,1,2}.$
  $endgroup$
  – Michael Rozenberg
  Jan 21 at 5:54
  

  
  
  
  

add a comment | 

2

$begingroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, why is {⊘} ⊂ {⊘,1,2} ?
  $endgroup$
  – J.S
  Jan 21 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
  $endgroup$
  – Anik Bhowmick
  Jan 21 at 5:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.S Because $oslashin{oslash,1,2}.$
  $endgroup$
  – Michael Rozenberg
  Jan 21 at 5:54
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

$endgroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

answered Jan 21 at 5:47

Michael RozenbergMichael Rozenberg

107k1894198

107k1894198

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, why is {⊘} ⊂ {⊘,1,2} ?
  $endgroup$
  – J.S
  Jan 21 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
  $endgroup$
  – Anik Bhowmick
  Jan 21 at 5:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.S Because $oslashin{oslash,1,2}.$
  $endgroup$
  – Michael Rozenberg
  Jan 21 at 5:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, why is {⊘} ⊂ {⊘,1,2} ?
  $endgroup$
  – J.S
  Jan 21 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
  $endgroup$
  – Anik Bhowmick
  Jan 21 at 5:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.S Because $oslashin{oslash,1,2}.$
  $endgroup$
  – Michael Rozenberg
  Jan 21 at 5:54
  

  
  
  
  

$begingroup$
Sorry, why is {⊘} ⊂ {⊘,1,2} ?
$endgroup$
– J.S
Jan 21 at 5:53

$begingroup$
Sorry, why is {⊘} ⊂ {⊘,1,2} ?
$endgroup$
– J.S
Jan 21 at 5:53

$begingroup$
Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
$endgroup$
– Anik Bhowmick
Jan 21 at 5:54

$begingroup$
Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
$endgroup$
– Anik Bhowmick
Jan 21 at 5:54

$begingroup$
@J.S Because $oslashin{oslash,1,2}.$
$endgroup$
– Michael Rozenberg
Jan 21 at 5:54

$begingroup$
@J.S Because $oslashin{oslash,1,2}.$
$endgroup$
– Michael Rozenberg
Jan 21 at 5:54

add a comment | 

0

$begingroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

$endgroup$

add a comment | 

0

$begingroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

$endgroup$

add a comment | 

0

0

0

$begingroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

$endgroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

answered Jan 21 at 5:47

Anik BhowmickAnik Bhowmick

583417

583417

add a comment | 

add a comment | 

0

$begingroup$

$neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.

Consider the case $A=phi.$

Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

$endgroup$

add a comment | 

0

$begingroup$

$neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.

Consider the case $A=phi.$

Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

$endgroup$

add a comment | 

0

0

0

$begingroup$

$neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.

Consider the case $A=phi.$

Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

$endgroup$

$neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.

Consider the case $A=phi.$

Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

answered Jan 21 at 6:59

DanielWainfleetDanielWainfleet

35.3k31648

35.3k31648

add a comment | 

add a comment | 

Thanks for contributing an answer to Mathematics Stack Exchange!

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But avoid …

• Asking for help, clarification, or responding to other answers.


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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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