Subset of a set that is in another set
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If we have a set, say
$$ A = emptyset, X $$
and we know that $B subset X$, does that imply $B subset A$?
elementary-set-theory
asked Sep 1 at 1:54
ketchup
645
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If we have a set, say
$$ A = emptyset, X $$
and we know that $B subset X$, does that imply $B subset A$?
elementary-set-theory
asked Sep 1 at 1:54
ketchup
645
3
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we have a set, say
$$ A = emptyset, X $$
and we know that $B subset X$, does that imply $B subset A$?
elementary-set-theory
asked Sep 1 at 1:54
ketchup
645
If we have a set, say
$$ A = emptyset, X $$
and we know that $B subset X$, does that imply $B subset A$?
elementary-set-theory
elementary-set-theory
asked Sep 1 at 1:54
ketchup
645
asked Sep 1 at 1:54
ketchup
645
asked Sep 1 at 1:54
ketchup
645
asked Sep 1 at 1:54
ketchup
645
asked Sep 1 at 1:54
ketchup
645
645
3
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22
add a comment |Â
3
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22
3
3
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22
add a comment |Â
1 Answer
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Answer -
No, it doesn't.
A Definition -
- Subset - A set, $mathbbY$ is said to be a subset of another set, $mathbbZ$, ( denoted as $mathbbY subset mathbbZ$ ) iff, $forall xin mathbbY$, $xin mathbbZ$.
Illustrative example -
Let, $mathbbX=1,2$ and $mathbbB=1$.
Now, note that $mathbbB$ is clearly a subset of $mathbbX$.
But, $mathbbA := phi, 1,2 = ,1,2$. Clearly, $1in mathbbB$, but $1not in mathbbA $. Thus, $mathbbB$ is not a subset of $mathbbA$.
General Explanation -
Here, your set $mathbbB$ is a subset of $mathbbX$. Now, if $mathbbX$ were to be a subset of $mathbbA$, then $mathbbB$ would clearly be a subset of $mathbbA$, I.e.$mathbbB subset mathbbX$ and $mathbbX subset mathbbA Rightarrow mathbbB subset mathbbA$.
But here, $mathbbX $ is merely an element of $mathbbA$. This means that the elements of $mathbbX$ are not elements of $mathbbA$ but are elements of a set which is an element of $mathbbA$.
Similarly, since $ mathbbB subset mathbbX $, we have the fact that it's elements are also not elements of $mathbbA$.
Thus, $mathbbB$ is not subset of $mathbbA$.
answered Sep 1 at 3:11
Devashish Kaushik
29714
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Answer -
No, it doesn't.
A Definition -
- Subset - A set, $mathbbY$ is said to be a subset of another set, $mathbbZ$, ( denoted as $mathbbY subset mathbbZ$ ) iff, $forall xin mathbbY$, $xin mathbbZ$.
Illustrative example -
Let, $mathbbX=1,2$ and $mathbbB=1$.
Now, note that $mathbbB$ is clearly a subset of $mathbbX$.
But, $mathbbA := phi, 1,2 = ,1,2$. Clearly, $1in mathbbB$, but $1not in mathbbA $. Thus, $mathbbB$ is not a subset of $mathbbA$.
General Explanation -
Here, your set $mathbbB$ is a subset of $mathbbX$. Now, if $mathbbX$ were to be a subset of $mathbbA$, then $mathbbB$ would clearly be a subset of $mathbbA$, I.e.$mathbbB subset mathbbX$ and $mathbbX subset mathbbA Rightarrow mathbbB subset mathbbA$.
But here, $mathbbX $ is merely an element of $mathbbA$. This means that the elements of $mathbbX$ are not elements of $mathbbA$ but are elements of a set which is an element of $mathbbA$.
Similarly, since $ mathbbB subset mathbbX $, we have the fact that it's elements are also not elements of $mathbbA$.
Thus, $mathbbB$ is not subset of $mathbbA$.
answered Sep 1 at 3:11
Devashish Kaushik
29714
add a comment |Â
up vote
0
down vote
Answer -
No, it doesn't.
A Definition -
- Subset - A set, $mathbbY$ is said to be a subset of another set, $mathbbZ$, ( denoted as $mathbbY subset mathbbZ$ ) iff, $forall xin mathbbY$, $xin mathbbZ$.
Illustrative example -
Let, $mathbbX=1,2$ and $mathbbB=1$.
Now, note that $mathbbB$ is clearly a subset of $mathbbX$.
But, $mathbbA := phi, 1,2 = ,1,2$. Clearly, $1in mathbbB$, but $1not in mathbbA $. Thus, $mathbbB$ is not a subset of $mathbbA$.
General Explanation -
Here, your set $mathbbB$ is a subset of $mathbbX$. Now, if $mathbbX$ were to be a subset of $mathbbA$, then $mathbbB$ would clearly be a subset of $mathbbA$, I.e.$mathbbB subset mathbbX$ and $mathbbX subset mathbbA Rightarrow mathbbB subset mathbbA$.
But here, $mathbbX $ is merely an element of $mathbbA$. This means that the elements of $mathbbX$ are not elements of $mathbbA$ but are elements of a set which is an element of $mathbbA$.
Similarly, since $ mathbbB subset mathbbX $, we have the fact that it's elements are also not elements of $mathbbA$.
Thus, $mathbbB$ is not subset of $mathbbA$.
answered Sep 1 at 3:11
Devashish Kaushik
29714
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Answer -
No, it doesn't.
A Definition -
- Subset - A set, $mathbbY$ is said to be a subset of another set, $mathbbZ$, ( denoted as $mathbbY subset mathbbZ$ ) iff, $forall xin mathbbY$, $xin mathbbZ$.
Illustrative example -
Let, $mathbbX=1,2$ and $mathbbB=1$.
Now, note that $mathbbB$ is clearly a subset of $mathbbX$.
But, $mathbbA := phi, 1,2 = ,1,2$. Clearly, $1in mathbbB$, but $1not in mathbbA $. Thus, $mathbbB$ is not a subset of $mathbbA$.
General Explanation -
Here, your set $mathbbB$ is a subset of $mathbbX$. Now, if $mathbbX$ were to be a subset of $mathbbA$, then $mathbbB$ would clearly be a subset of $mathbbA$, I.e.$mathbbB subset mathbbX$ and $mathbbX subset mathbbA Rightarrow mathbbB subset mathbbA$.
But here, $mathbbX $ is merely an element of $mathbbA$. This means that the elements of $mathbbX$ are not elements of $mathbbA$ but are elements of a set which is an element of $mathbbA$.
Similarly, since $ mathbbB subset mathbbX $, we have the fact that it's elements are also not elements of $mathbbA$.
Thus, $mathbbB$ is not subset of $mathbbA$.
answered Sep 1 at 3:11
Devashish Kaushik
29714
Answer -
No, it doesn't.
A Definition -
- Subset - A set, $mathbbY$ is said to be a subset of another set, $mathbbZ$, ( denoted as $mathbbY subset mathbbZ$ ) iff, $forall xin mathbbY$, $xin mathbbZ$.
Illustrative example -
Let, $mathbbX=1,2$ and $mathbbB=1$.
Now, note that $mathbbB$ is clearly a subset of $mathbbX$.
But, $mathbbA := phi, 1,2 = ,1,2$. Clearly, $1in mathbbB$, but $1not in mathbbA $. Thus, $mathbbB$ is not a subset of $mathbbA$.
General Explanation -
Here, your set $mathbbB$ is a subset of $mathbbX$. Now, if $mathbbX$ were to be a subset of $mathbbA$, then $mathbbB$ would clearly be a subset of $mathbbA$, I.e.$mathbbB subset mathbbX$ and $mathbbX subset mathbbA Rightarrow mathbbB subset mathbbA$.
But here, $mathbbX $ is merely an element of $mathbbA$. This means that the elements of $mathbbX$ are not elements of $mathbbA$ but are elements of a set which is an element of $mathbbA$.
Similarly, since $ mathbbB subset mathbbX $, we have the fact that it's elements are also not elements of $mathbbA$.
Thus, $mathbbB$ is not subset of $mathbbA$.
answered Sep 1 at 3:11
Devashish Kaushik
29714
edited Sep 1 at 3:18
answered Sep 1 at 3:11
Devashish Kaushik
29714
answered Sep 1 at 3:11
Devashish Kaushik
29714
answered Sep 1 at 3:11
Devashish Kaushik
29714
29714
add a comment |Â
add a comment |Â
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3
No, in exactly the same way that $xin X$ does not imply that $xin A$. $A$ only contains two elements $emptyset$ and $X$. That these elements are sets themselves is irrelevant. The only subsets of $A$ are then $emptyset, emptyset, X$ and $emptyset,X$.
– JMoravitz
Sep 1 at 1:59
No. For example, set $X=1, 2, 3$ and $B=2, 3$. Then clearly $B subset X$, but it isn't true that $B subset A$ because the only two elements of $A$ are $varnothing$ and $X$, and neither is $B$.
– Rócherz
Sep 1 at 2:02
@JMoravitz Why are you posting a perfectly good answer as a comment?
– Arthur
Sep 1 at 3:01
No, in computer language terms that would be a type error.
– Henno Brandsma
Sep 1 at 6:22