Mixing
Let $A$ be a subset. If $A$ is non-empty, then $A$ has at least two subsets. Specifically, $emptyset$...
$begingroup$
Still new to proofs so walking through this step-by-step.
Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.
Proving the contrapositive.
If $A$ has less than two subsets, then $A$ is empty.
Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?
I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.
elementary-set-theory
asked Jan 7 at 21:24
RyanRyan
1356
$endgroup$
add a comment |
$begingroup$
Still new to proofs so walking through this step-by-step.
Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.
Proving the contrapositive.
If $A$ has less than two subsets, then $A$ is empty.
Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?
I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.
elementary-set-theory
asked Jan 7 at 21:24
RyanRyan
1356
$endgroup$
1
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
3
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26
add a comment |
$begingroup$
Still new to proofs so walking through this step-by-step.
Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.
Proving the contrapositive.
If $A$ has less than two subsets, then $A$ is empty.
Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?
I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.
elementary-set-theory
asked Jan 7 at 21:24
RyanRyan
1356
$endgroup$
Still new to proofs so walking through this step-by-step.
Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.
Proving the contrapositive.
If $A$ has less than two subsets, then $A$ is empty.
Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?
I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.
elementary-set-theory
elementary-set-theory
asked Jan 7 at 21:24
RyanRyan
1356
asked Jan 7 at 21:24
RyanRyan
1356
asked Jan 7 at 21:24
RyanRyan
1356
asked Jan 7 at 21:24
RyanRyan
1356
asked Jan 7 at 21:24
RyanRyan
1356
1356
1
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
3
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26
add a comment |
1
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
3
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26
1
1
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
3
3
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If A is empty, then A does not have two subsets.
Thus, if A has two subsets, A is not empty.
edited Jan 8 at 9:35
Math_QED
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If A is empty, then A does not have two subsets.
Thus, if A has two subsets, A is not empty.
edited Jan 8 at 9:35
Math_QED
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
add a comment |
$begingroup$
If A is empty, then A does not have two subsets.
Thus, if A has two subsets, A is not empty.
edited Jan 8 at 9:35
Math_QED
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
add a comment |
$begingroup$
If A is empty, then A does not have two subsets.
Thus, if A has two subsets, A is not empty.
edited Jan 8 at 9:35
Math_QED
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
$endgroup$
If A is empty, then A does not have two subsets.
Thus, if A has two subsets, A is not empty.
edited Jan 8 at 9:35
Math_QED
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
edited Jan 8 at 9:35
Math_QED
7,52831450
edited Jan 8 at 9:35
Math_QED
7,52831450
edited Jan 8 at 9:35
Math_QED
7,52831450
7,52831450
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
answered Jan 8 at 9:31
William ElliotWilliam Elliot
7,8502720
7,8502720
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
add a comment |
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30
add a comment |
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1
$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25
3
$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26