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Is the word “empty†in set theory different from the word “empty†in ordinary language? [duplicate]
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This question already has an answer here:
How is an empty set truly “empty�
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I skimmed over this question Is the empty set a subset of itself?, and I'm currently under the impression that it's a widespread belief that the empty set contains itself. But a contradiction seems to arise: if the empty set contains itself, then it's NOT empty. After all, if we call a set non-empty just because it contains the empty set, then why should we treat the empty set itself differently?
Then it occurred to me that maybe mathematicians define "empty" differently in set theory. Maybe by "empty set" they mean a "set that contains only itself", instead of a "set that contains absolutely nothing".
Is my surmise correct?
elementary-set-theory
edited 2 days ago
amWhy
190k26221433
asked 2 days ago
user161005
917
marked as duplicate by Christoph, Jendrik Stelzner, José Carlos Santos, Strants, Lord Shark the Unknown 2 days ago
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.
 |Â
show 7 more comments
up vote
4
down vote
favorite
This question already has an answer here:
How is an empty set truly “empty�
6 answers
I skimmed over this question Is the empty set a subset of itself?, and I'm currently under the impression that it's a widespread belief that the empty set contains itself. But a contradiction seems to arise: if the empty set contains itself, then it's NOT empty. After all, if we call a set non-empty just because it contains the empty set, then why should we treat the empty set itself differently?
Then it occurred to me that maybe mathematicians define "empty" differently in set theory. Maybe by "empty set" they mean a "set that contains only itself", instead of a "set that contains absolutely nothing".
Is my surmise correct?
elementary-set-theory
edited 2 days ago
amWhy
190k26221433
asked 2 days ago
user161005
917
marked as duplicate by Christoph, Jendrik Stelzner, José Carlos Santos, Strants, Lord Shark the Unknown 2 days ago
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.
13
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
5
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
2
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
1
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago
 |Â
show 7 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
How is an empty set truly “empty�
6 answers
I skimmed over this question Is the empty set a subset of itself?, and I'm currently under the impression that it's a widespread belief that the empty set contains itself. But a contradiction seems to arise: if the empty set contains itself, then it's NOT empty. After all, if we call a set non-empty just because it contains the empty set, then why should we treat the empty set itself differently?
Then it occurred to me that maybe mathematicians define "empty" differently in set theory. Maybe by "empty set" they mean a "set that contains only itself", instead of a "set that contains absolutely nothing".
Is my surmise correct?
elementary-set-theory
edited 2 days ago
amWhy
190k26221433
asked 2 days ago
user161005
917
This question already has an answer here:
How is an empty set truly “empty�
6 answers
I skimmed over this question Is the empty set a subset of itself?, and I'm currently under the impression that it's a widespread belief that the empty set contains itself. But a contradiction seems to arise: if the empty set contains itself, then it's NOT empty. After all, if we call a set non-empty just because it contains the empty set, then why should we treat the empty set itself differently?
Then it occurred to me that maybe mathematicians define "empty" differently in set theory. Maybe by "empty set" they mean a "set that contains only itself", instead of a "set that contains absolutely nothing".
Is my surmise correct?
This question already has an answer here:
How is an empty set truly “empty�
6 answers
elementary-set-theory
elementary-set-theory
edited 2 days ago
amWhy
190k26221433
asked 2 days ago
user161005
917
edited 2 days ago
amWhy
190k26221433
asked 2 days ago
user161005
917
edited 2 days ago
amWhy
190k26221433
edited 2 days ago
amWhy
190k26221433
edited 2 days ago
amWhy
190k26221433
190k26221433
asked 2 days ago
user161005
917
asked 2 days ago
user161005
917
asked 2 days ago
user161005
917
917
marked as duplicate by Christoph, Jendrik Stelzner, José Carlos Santos, Strants, Lord Shark the Unknown 2 days ago
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 Christoph, Jendrik Stelzner, José Carlos Santos, Strants, Lord Shark the Unknown 2 days ago
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.
13
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
5
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
2
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
1
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago
 |Â
show 7 more comments
13
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
5
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
2
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
1
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago
13
13
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
5
5
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
2
2
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
1
1
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago
 |Â
show 7 more comments
3 Answers
3
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up vote
19
down vote
accepted
I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)
However, $varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.
To see the distinction in non-empty sets, $1$ is a subset of $1,2$ but $1,2$ does not contain $1$. It does contain $1$, but that is not the same thing.
On the usage of "contains", wikipedia says:
The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.
answered 2 days ago
Especially Lime
19.7k22353
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
 |Â
show 2 more comments
up vote
9
down vote
The problem here is you are using the word "contains" for two different things:
- When $x$ is an element of a set $A$, we write $xin A$ and sometimes say "$A$ contains $x$".
- When $B$ is a subset of $A$, we write $Bsubseteq A$ and also sometimes say "$A$ contains $B$".
This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.
With the empty set, it is always false that $xinvarnothing$, i.e., the empty set does not have elements, it is empty.
However, it has a subset: $varnothingsubseteqvarnothing$ is true. Why is that? Well, $Bsubseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $varnothingsubseteq X$ is true for any set $X$, since there are no elements in $varnothing$ that have to be contained in $X$ at all.
answered 2 days ago
Christoph
10.9k1240
add a comment |Â
up vote
3
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$Asubseteq B$ is not the same as $Ain B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=,A,dots,$.
Therefore it is correct to say $emptysetsubseteqemptyset$ but not $emptysetinemptyset$ since the empty set has no elements.
answered 2 days ago
EuklidAlexandria
2739
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
19
down vote
accepted
I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)
However, $varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.
To see the distinction in non-empty sets, $1$ is a subset of $1,2$ but $1,2$ does not contain $1$. It does contain $1$, but that is not the same thing.
On the usage of "contains", wikipedia says:
The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.
answered 2 days ago
Especially Lime
19.7k22353
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
 |Â
show 2 more comments
up vote
19
down vote
accepted
I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)
However, $varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.
To see the distinction in non-empty sets, $1$ is a subset of $1,2$ but $1,2$ does not contain $1$. It does contain $1$, but that is not the same thing.
On the usage of "contains", wikipedia says:
The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.
answered 2 days ago
Especially Lime
19.7k22353
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
 |Â
show 2 more comments
up vote
19
down vote
accepted
up vote
19
down vote
accepted
I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)
However, $varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.
To see the distinction in non-empty sets, $1$ is a subset of $1,2$ but $1,2$ does not contain $1$. It does contain $1$, but that is not the same thing.
On the usage of "contains", wikipedia says:
The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.
answered 2 days ago
Especially Lime
19.7k22353
I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)
However, $varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.
To see the distinction in non-empty sets, $1$ is a subset of $1,2$ but $1,2$ does not contain $1$. It does contain $1$, but that is not the same thing.
On the usage of "contains", wikipedia says:
The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.
answered 2 days ago
Especially Lime
19.7k22353
edited 2 days ago
answered 2 days ago
Especially Lime
19.7k22353
answered 2 days ago
Especially Lime
19.7k22353
answered 2 days ago
Especially Lime
19.7k22353
19.7k22353
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
 |Â
show 2 more comments
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
"It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set 1 as element? And would be it possible for 1 to be an element without being a subset?
– user161005
2 days ago
2
2
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
Sure, the power set (set of all subsets) of $1,2$ is $P=1,2,1,2,varnothing$. So $1in P$, but $1notin P$, meaning that $1notsubset P$.
– Especially Lime
2 days ago
2
2
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
The opening paragraph has the contains relationship backwards.
– jaxad0127
2 days ago
1
1
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
@user161005: By definition something is an element of $1,2$ if and only if it equals $1$ or it equals $2$. We cannot have $1=1$ due to the Axiom of Regularity. Whether $1=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $1$, but for the last many decades it has been nearly universal to define $2$ to mean $0,1$. In that case $1=2$ is false because $0in 2$ but $0notin1$. Therefore, $1notin1,2$.
– Henning Makholm
2 days ago
1
1
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
@HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $emptyset subseteq emptyset$?"
– Vincent
2 days ago
 |Â
show 2 more comments
up vote
9
down vote
The problem here is you are using the word "contains" for two different things:
- When $x$ is an element of a set $A$, we write $xin A$ and sometimes say "$A$ contains $x$".
- When $B$ is a subset of $A$, we write $Bsubseteq A$ and also sometimes say "$A$ contains $B$".
This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.
With the empty set, it is always false that $xinvarnothing$, i.e., the empty set does not have elements, it is empty.
However, it has a subset: $varnothingsubseteqvarnothing$ is true. Why is that? Well, $Bsubseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $varnothingsubseteq X$ is true for any set $X$, since there are no elements in $varnothing$ that have to be contained in $X$ at all.
answered 2 days ago
Christoph
10.9k1240
add a comment |Â
up vote
9
down vote
The problem here is you are using the word "contains" for two different things:
- When $x$ is an element of a set $A$, we write $xin A$ and sometimes say "$A$ contains $x$".
- When $B$ is a subset of $A$, we write $Bsubseteq A$ and also sometimes say "$A$ contains $B$".
This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.
With the empty set, it is always false that $xinvarnothing$, i.e., the empty set does not have elements, it is empty.
However, it has a subset: $varnothingsubseteqvarnothing$ is true. Why is that? Well, $Bsubseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $varnothingsubseteq X$ is true for any set $X$, since there are no elements in $varnothing$ that have to be contained in $X$ at all.
answered 2 days ago
Christoph
10.9k1240
add a comment |Â
up vote
9
down vote
up vote
9
down vote
The problem here is you are using the word "contains" for two different things:
- When $x$ is an element of a set $A$, we write $xin A$ and sometimes say "$A$ contains $x$".
- When $B$ is a subset of $A$, we write $Bsubseteq A$ and also sometimes say "$A$ contains $B$".
This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.
With the empty set, it is always false that $xinvarnothing$, i.e., the empty set does not have elements, it is empty.
However, it has a subset: $varnothingsubseteqvarnothing$ is true. Why is that? Well, $Bsubseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $varnothingsubseteq X$ is true for any set $X$, since there are no elements in $varnothing$ that have to be contained in $X$ at all.
answered 2 days ago
Christoph
10.9k1240
The problem here is you are using the word "contains" for two different things:
- When $x$ is an element of a set $A$, we write $xin A$ and sometimes say "$A$ contains $x$".
- When $B$ is a subset of $A$, we write $Bsubseteq A$ and also sometimes say "$A$ contains $B$".
This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.
With the empty set, it is always false that $xinvarnothing$, i.e., the empty set does not have elements, it is empty.
However, it has a subset: $varnothingsubseteqvarnothing$ is true. Why is that? Well, $Bsubseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $varnothingsubseteq X$ is true for any set $X$, since there are no elements in $varnothing$ that have to be contained in $X$ at all.
answered 2 days ago
Christoph
10.9k1240
answered 2 days ago
Christoph
10.9k1240
answered 2 days ago
Christoph
10.9k1240
answered 2 days ago
Christoph
10.9k1240
10.9k1240
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up vote
3
down vote
$Asubseteq B$ is not the same as $Ain B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=,A,dots,$.
Therefore it is correct to say $emptysetsubseteqemptyset$ but not $emptysetinemptyset$ since the empty set has no elements.
answered 2 days ago
EuklidAlexandria
2739
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
add a comment |Â
up vote
3
down vote
$Asubseteq B$ is not the same as $Ain B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=,A,dots,$.
Therefore it is correct to say $emptysetsubseteqemptyset$ but not $emptysetinemptyset$ since the empty set has no elements.
answered 2 days ago
EuklidAlexandria
2739
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$Asubseteq B$ is not the same as $Ain B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=,A,dots,$.
Therefore it is correct to say $emptysetsubseteqemptyset$ but not $emptysetinemptyset$ since the empty set has no elements.
answered 2 days ago
EuklidAlexandria
2739
$Asubseteq B$ is not the same as $Ain B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=,A,dots,$.
Therefore it is correct to say $emptysetsubseteqemptyset$ but not $emptysetinemptyset$ since the empty set has no elements.
answered 2 days ago
EuklidAlexandria
2739
edited 2 days ago
answered 2 days ago
EuklidAlexandria
2739
answered 2 days ago
EuklidAlexandria
2739
answered 2 days ago
EuklidAlexandria
2739
2739
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
add a comment |Â
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
2
2
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
Along with the ambiguity of "contains", I really wish people would stop using $subset$ to mean $subseteq$. Nobody would ever write $<$ to mean $leq$ so why do we do it with sets?
– David Richerby
2 days ago
2
2
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
@DavidRicherby Thank you. I have changed it. In university we have used $subset$ and $subsetneq$ so far.
– EuklidAlexandria
2 days ago
add a comment |Â
13
"Contains" is an ambiguous word that people should stop using with regard to sets.
– Malice Vidrine
2 days ago
5
"contained" $in$ is different (in set theory) from "contained" $subseteq$.
– Mauro ALLEGRANZA
2 days ago
@MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset.
– user161005
2 days ago
2
NO. The basic relation between sets is : $A in B$. Empty set is defined as the set satisfying the formula : $exists E forall x lnot (x in E)$.
– Mauro ALLEGRANZA
2 days ago
1
Subset is defined as : $A subseteq B leftrightarrow forall x (x in A to x in B)$.
– Mauro ALLEGRANZA
2 days ago