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Is $∅$ proper subset of $∅$? [duplicate]
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If null set is an element of a set then will it belongs to set or subset? [closed]
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I get that $∅$ is subset of every set thus $∅ ⊆ ∅$.
However, I'm not sure if $∅ ⊂ ∅$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $∅$ have an element that $∅$ doesn't have?
elementary-set-theory
edited Aug 2 at 8:34
Asaf Karagila
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asked Aug 2 at 6:30
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marked as duplicate by M. Winter, Mostafa Ayaz, amWhy elementary-set-theory
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Aug 2 at 10:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
If null set is an element of a set then will it belongs to set or subset? [closed]
4 answers
I get that $∅$ is subset of every set thus $∅ ⊆ ∅$.
However, I'm not sure if $∅ ⊂ ∅$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $∅$ have an element that $∅$ doesn't have?
elementary-set-theory
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
asked Aug 2 at 6:30
Jae
232
marked as duplicate by M. Winter, Mostafa Ayaz, amWhy elementary-set-theory
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Aug 2 at 10:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
1
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
2
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29
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up vote
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This question already has an answer here:
If null set is an element of a set then will it belongs to set or subset? [closed]
4 answers
I get that $∅$ is subset of every set thus $∅ ⊆ ∅$.
However, I'm not sure if $∅ ⊂ ∅$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $∅$ have an element that $∅$ doesn't have?
elementary-set-theory
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
asked Aug 2 at 6:30
Jae
232
This question already has an answer here:
If null set is an element of a set then will it belongs to set or subset? [closed]
4 answers
I get that $∅$ is subset of every set thus $∅ ⊆ ∅$.
However, I'm not sure if $∅ ⊂ ∅$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $∅$ have an element that $∅$ doesn't have?
This question already has an answer here:
If null set is an element of a set then will it belongs to set or subset? [closed]
4 answers
elementary-set-theory
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
asked Aug 2 at 6:30
Jae
232
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
edited Aug 2 at 8:34
Asaf Karagila
291k31401731
291k31401731
asked Aug 2 at 6:30
Jae
232
asked Aug 2 at 6:30
Jae
232
asked Aug 2 at 6:30
Jae
232
232
marked as duplicate by M. Winter, Mostafa Ayaz, amWhy elementary-set-theory
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Aug 2 at 10:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by M. Winter, Mostafa Ayaz, amWhy elementary-set-theory
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Aug 2 at 10:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
1
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
2
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29
add a comment |Â
1
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
1
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
2
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29
1
1
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
1
1
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
2
2
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29
add a comment |Â
3 Answers
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The set $emptyset$ contains the element $emptyset$. The empty set contains no elements, thus the containment is proper i.e $emptysetsubsetneq emptyset$.
answered Aug 2 at 6:34
Alvin Lepik
2,035819
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4
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Yes. $emptysetsubsetneqemptyset$ becasue $emptysetinemptyset$ and $emptysetnotinemptyset$. More generally, empty set is proper subset of every non-empty set.
answered Aug 2 at 7:13
Le Anh Dung
728317
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Let A=∅ be a set, then ∅∈A.
But always remember, ∅ ≠∅, the former is an empty set and the latter is the element of a set.
Hence set A contains one element (∅). And as per your definition,From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
∅ ⊂ ∅, as ∅ is an empty set containing no element and ∅ is a set containing one element (∅).
answered Aug 2 at 7:06
Ashwani Bhat
1
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The set $emptyset$ contains the element $emptyset$. The empty set contains no elements, thus the containment is proper i.e $emptysetsubsetneq emptyset$.
answered Aug 2 at 6:34
Alvin Lepik
2,035819
add a comment |Â
up vote
6
down vote
accepted
The set $emptyset$ contains the element $emptyset$. The empty set contains no elements, thus the containment is proper i.e $emptysetsubsetneq emptyset$.
answered Aug 2 at 6:34
Alvin Lepik
2,035819
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The set $emptyset$ contains the element $emptyset$. The empty set contains no elements, thus the containment is proper i.e $emptysetsubsetneq emptyset$.
answered Aug 2 at 6:34
Alvin Lepik
2,035819
The set $emptyset$ contains the element $emptyset$. The empty set contains no elements, thus the containment is proper i.e $emptysetsubsetneq emptyset$.
answered Aug 2 at 6:34
Alvin Lepik
2,035819
answered Aug 2 at 6:34
Alvin Lepik
2,035819
answered Aug 2 at 6:34
Alvin Lepik
2,035819
answered Aug 2 at 6:34
Alvin Lepik
2,035819
2,035819
add a comment |Â
add a comment |Â
up vote
4
down vote
Yes. $emptysetsubsetneqemptyset$ becasue $emptysetinemptyset$ and $emptysetnotinemptyset$. More generally, empty set is proper subset of every non-empty set.
answered Aug 2 at 7:13
Le Anh Dung
728317
add a comment |Â
up vote
4
down vote
Yes. $emptysetsubsetneqemptyset$ becasue $emptysetinemptyset$ and $emptysetnotinemptyset$. More generally, empty set is proper subset of every non-empty set.
answered Aug 2 at 7:13
Le Anh Dung
728317
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Yes. $emptysetsubsetneqemptyset$ becasue $emptysetinemptyset$ and $emptysetnotinemptyset$. More generally, empty set is proper subset of every non-empty set.
answered Aug 2 at 7:13
Le Anh Dung
728317
Yes. $emptysetsubsetneqemptyset$ becasue $emptysetinemptyset$ and $emptysetnotinemptyset$. More generally, empty set is proper subset of every non-empty set.
answered Aug 2 at 7:13
Le Anh Dung
728317
edited Aug 2 at 7:27
answered Aug 2 at 7:13
Le Anh Dung
728317
answered Aug 2 at 7:13
Le Anh Dung
728317
answered Aug 2 at 7:13
Le Anh Dung
728317
728317
add a comment |Â
add a comment |Â
up vote
0
down vote
Let A=∅ be a set, then ∅∈A.
But always remember, ∅ ≠∅, the former is an empty set and the latter is the element of a set.
Hence set A contains one element (∅). And as per your definition,From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
∅ ⊂ ∅, as ∅ is an empty set containing no element and ∅ is a set containing one element (∅).
answered Aug 2 at 7:06
Ashwani Bhat
1
add a comment |Â
up vote
0
down vote
Let A=∅ be a set, then ∅∈A.
But always remember, ∅ ≠∅, the former is an empty set and the latter is the element of a set.
Hence set A contains one element (∅). And as per your definition,From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
∅ ⊂ ∅, as ∅ is an empty set containing no element and ∅ is a set containing one element (∅).
answered Aug 2 at 7:06
Ashwani Bhat
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let A=∅ be a set, then ∅∈A.
But always remember, ∅ ≠∅, the former is an empty set and the latter is the element of a set.
Hence set A contains one element (∅). And as per your definition,From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
∅ ⊂ ∅, as ∅ is an empty set containing no element and ∅ is a set containing one element (∅).
answered Aug 2 at 7:06
Ashwani Bhat
1
Let A=∅ be a set, then ∅∈A.
But always remember, ∅ ≠∅, the former is an empty set and the latter is the element of a set.
Hence set A contains one element (∅). And as per your definition,From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
∅ ⊂ ∅, as ∅ is an empty set containing no element and ∅ is a set containing one element (∅).
answered Aug 2 at 7:06
Ashwani Bhat
1
answered Aug 2 at 7:06
Ashwani Bhat
1
answered Aug 2 at 7:06
Ashwani Bhat
1
answered Aug 2 at 7:06
Ashwani Bhat
1
1
add a comment |Â
add a comment |Â
1
"From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one". You can also phrase it more directy as "$A$ is a proper subset of $B$ if and only if $Asubseteq B$ and $Ane B$".
– Saucy O'Path
Aug 2 at 6:48
1
The empty set is subset of every set; thus it is a proper subset of every not-empty set:
– Mauro ALLEGRANZA
Aug 2 at 7:08
2
$varnothing$ is an element of $varnothing$ which is not an element of $varnothing$. I explained this in my answer over here.
– M. Winter
Aug 2 at 8:29