Mixing
Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.
$begingroup$
Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.
I think this is false because of the counterexample
$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$
but I am not sure if I am right.
discrete-mathematics elementary-set-theory
edited Jan 11 at 20:16
6005
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
$endgroup$
add a comment |
$begingroup$
Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.
I think this is false because of the counterexample
$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$
but I am not sure if I am right.
discrete-mathematics elementary-set-theory
edited Jan 11 at 20:16
6005
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
$endgroup$
9
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
4
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58
add a comment |
$begingroup$
Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.
I think this is false because of the counterexample
$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$
but I am not sure if I am right.
discrete-mathematics elementary-set-theory
edited Jan 11 at 20:16
6005
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
$endgroup$
Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.
I think this is false because of the counterexample
$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$
but I am not sure if I am right.
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Jan 11 at 20:16
6005
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
edited Jan 11 at 20:16
6005
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
edited Jan 11 at 20:16
6005
36.2k751125
edited Jan 11 at 20:16
6005
36.2k751125
edited Jan 11 at 20:16
6005
36.2k751125
36.2k751125
asked Jan 11 at 19:53
hkdhkd
411
asked Jan 11 at 19:53
hkdhkd
411
asked Jan 11 at 19:53
hkdhkd
411
411
9
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
4
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58
add a comment |
9
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
4
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58
9
9
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
4
4
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nice counterexample, you are correct. Let us check each condition:
$A$ is a subset of $B$
That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.
$B$ belongs to $C$
That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.
then $A$ belongs to $C$
This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.
So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.
answered Jan 11 at 20:15
60056005
36.2k751125
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Nice counterexample, you are correct. Let us check each condition:
$A$ is a subset of $B$
That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.
$B$ belongs to $C$
That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.
then $A$ belongs to $C$
This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.
So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.
answered Jan 11 at 20:15
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
Nice counterexample, you are correct. Let us check each condition:
$A$ is a subset of $B$
That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.
$B$ belongs to $C$
That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.
then $A$ belongs to $C$
This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.
So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.
answered Jan 11 at 20:15
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
Nice counterexample, you are correct. Let us check each condition:
$A$ is a subset of $B$
That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.
$B$ belongs to $C$
That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.
then $A$ belongs to $C$
This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.
So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.
answered Jan 11 at 20:15
60056005
36.2k751125
$endgroup$
Nice counterexample, you are correct. Let us check each condition:
$A$ is a subset of $B$
That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.
$B$ belongs to $C$
That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.
then $A$ belongs to $C$
This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.
So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.
answered Jan 11 at 20:15
60056005
36.2k751125
answered Jan 11 at 20:15
60056005
36.2k751125
answered Jan 11 at 20:15
60056005
36.2k751125
answered Jan 11 at 20:15
60056005
36.2k751125
36.2k751125
add a comment |
add a comment |
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9
$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54
4
$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58