Theorems in the form of “if and only if” such that the proof of one direction is extremely EASY to prove and the other one is extremely HARD [closed]
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The 'if and only if' formulation should be as natural as possible.
Thanks in advance.
Edit: I know this question is somewhat ill-posed. A better one: Theorems in the form of “if and only if” such that the proofs of BOTH directions are nontrivial
soft-question big-list
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
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closed as too broad by BlueRaja - Danny Pflughoeft, Arnaud D., Cesareo, Mees de Vries, Carl Mummert Jan 11 at 12:33
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The 'if and only if' formulation should be as natural as possible.
Thanks in advance.
Edit: I know this question is somewhat ill-posed. A better one: Theorems in the form of “if and only if” such that the proofs of BOTH directions are nontrivial
soft-question big-list
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
$endgroup$
closed as too broad by BlueRaja - Danny Pflughoeft, Arnaud D., Cesareo, Mees de Vries, Carl Mummert Jan 11 at 12:33
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
$endgroup$
– Blue
Jan 11 at 5:21
2
$begingroup$
That's a good metaphor!
$endgroup$
– YuiTo Cheng
Jan 11 at 5:23
11
$begingroup$
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
$endgroup$
– bof
Jan 11 at 5:39
$begingroup$
@bof done, see math.stackexchange.com/questions/3069590/…
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– YuiTo Cheng
Jan 11 at 7:57
3
$begingroup$
You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
$endgroup$
– Nij
Jan 11 at 9:57
|
show 3 more comments
$begingroup$
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The 'if and only if' formulation should be as natural as possible.
Thanks in advance.
Edit: I know this question is somewhat ill-posed. A better one: Theorems in the form of “if and only if” such that the proofs of BOTH directions are nontrivial
soft-question big-list
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
$endgroup$
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The 'if and only if' formulation should be as natural as possible.
Thanks in advance.
Edit: I know this question is somewhat ill-posed. A better one: Theorems in the form of “if and only if” such that the proofs of BOTH directions are nontrivial
soft-question big-list
soft-question big-list
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
edited Jan 11 at 12:59
YuiTo Cheng
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
asked Jan 11 at 5:07
YuiTo ChengYuiTo Cheng
350117
350117
closed as too broad by BlueRaja - Danny Pflughoeft, Arnaud D., Cesareo, Mees de Vries, Carl Mummert Jan 11 at 12:33
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by BlueRaja - Danny Pflughoeft, Arnaud D., Cesareo, Mees de Vries, Carl Mummert Jan 11 at 12:33
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
$endgroup$
– Blue
Jan 11 at 5:21
2
$begingroup$
That's a good metaphor!
$endgroup$
– YuiTo Cheng
Jan 11 at 5:23
11
$begingroup$
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
$endgroup$
– bof
Jan 11 at 5:39
$begingroup$
@bof done, see math.stackexchange.com/questions/3069590/…
$endgroup$
– YuiTo Cheng
Jan 11 at 7:57
3
$begingroup$
You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
$endgroup$
– Nij
Jan 11 at 9:57
|
show 3 more comments
3
$begingroup$
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
$endgroup$
– Blue
Jan 11 at 5:21
2
$begingroup$
That's a good metaphor!
$endgroup$
– YuiTo Cheng
Jan 11 at 5:23
11
$begingroup$
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
$endgroup$
– bof
Jan 11 at 5:39
$begingroup$
@bof done, see math.stackexchange.com/questions/3069590/…
$endgroup$
– YuiTo Cheng
Jan 11 at 7:57
3
$begingroup$
You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
$endgroup$
– Nij
Jan 11 at 9:57
3
3
$begingroup$
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
$endgroup$
– Blue
Jan 11 at 5:21
$begingroup$
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
$endgroup$
– Blue
Jan 11 at 5:21
2
2
$begingroup$
That's a good metaphor!
$endgroup$
– YuiTo Cheng
Jan 11 at 5:23
$begingroup$
That's a good metaphor!
$endgroup$
– YuiTo Cheng
Jan 11 at 5:23
11
11
$begingroup$
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
$endgroup$
– bof
Jan 11 at 5:39
$begingroup$
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
$endgroup$
– bof
Jan 11 at 5:39
$begingroup$
@bof done, see math.stackexchange.com/questions/3069590/…
$endgroup$
– YuiTo Cheng
Jan 11 at 7:57
$begingroup$
@bof done, see math.stackexchange.com/questions/3069590/…
$endgroup$
– YuiTo Cheng
Jan 11 at 7:57
3
3
$begingroup$
You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
$endgroup$
– Nij
Jan 11 at 9:57
$begingroup$
You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
$endgroup$
– Nij
Jan 11 at 9:57
|
show 3 more comments
9 Answers
9
active
oldest
votes
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"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered Jan 11 at 7:36
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
$endgroup$
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
add a comment |
$begingroup$
All planar graphs are $n$-colorable iff $nge4$.
answered Jan 11 at 7:38
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
$endgroup$
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
add a comment |
$begingroup$
The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_3,3$ or $K_5$.
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
$endgroup$
add a comment |
$begingroup$
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
$endgroup$
2
$begingroup$
Assuming they're all non-empty.
$endgroup$
– Arnaud D.
Jan 11 at 9:14
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Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
$endgroup$
– Marc Paul
Jan 14 at 22:59
add a comment |
$begingroup$
An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
$endgroup$
6
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
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– YuiTo Cheng
Jan 11 at 9:20
add a comment |
$begingroup$
$mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
answered Jan 11 at 12:30
guest9366710guest9366710
111
$endgroup$
add a comment |
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
$endgroup$
add a comment |
$begingroup$
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
$endgroup$
add a comment |
$begingroup$
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
$endgroup$
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
answered Jan 11 at 5:23
Alvin JinAlvin Jin
2,2221019
2,2221019
add a comment |
add a comment |
$begingroup$
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered Jan 11 at 7:36
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered Jan 11 at 7:36
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered Jan 11 at 7:36
bofbof
51.3k457120
$endgroup$
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered Jan 11 at 7:36
bofbof
51.3k457120
answered Jan 11 at 7:36
bofbof
51.3k457120
answered Jan 11 at 7:36
bofbof
51.3k457120
answered Jan 11 at 7:36
bofbof
51.3k457120
51.3k457120
add a comment |
add a comment |
$begingroup$
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
$endgroup$
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
add a comment |
$begingroup$
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
$endgroup$
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
add a comment |
$begingroup$
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
$endgroup$
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
answered Jan 11 at 7:09
lonza leggieralonza leggiera
4615
4615
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
add a comment |
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
2
2
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
$begingroup$
The Steiner–Lehmus theorem.
$endgroup$
– Rosie F
Jan 11 at 9:08
add a comment |
$begingroup$
All planar graphs are $n$-colorable iff $nge4$.
answered Jan 11 at 7:38
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
All planar graphs are $n$-colorable iff $nge4$.
answered Jan 11 at 7:38
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
All planar graphs are $n$-colorable iff $nge4$.
answered Jan 11 at 7:38
bofbof
51.3k457120
$endgroup$
All planar graphs are $n$-colorable iff $nge4$.
answered Jan 11 at 7:38
bofbof
51.3k457120
answered Jan 11 at 7:38
bofbof
51.3k457120
answered Jan 11 at 7:38
bofbof
51.3k457120
answered Jan 11 at 7:38
bofbof
51.3k457120
51.3k457120
add a comment |
add a comment |
$begingroup$
Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
$endgroup$
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
add a comment |
$begingroup$
Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
$endgroup$
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
add a comment |
$begingroup$
Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
$endgroup$
Integers $a * b = 944871836856449473$ and $b > a > 1$
iff
$a = 961748941$ and $b = 982451653$
If is trivial multiplication. Only if requires large prime factorization.
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
edited Jan 11 at 8:28
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
answered Jan 11 at 8:24
MooseBoysMooseBoys
1895
1895
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
add a comment |
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
11
11
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
$begingroup$
The hard direction only requires primality testing, not factorization.
$endgroup$
– bof
Jan 11 at 9:10
add a comment |
$begingroup$
The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_3,3$ or $K_5$.
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
$endgroup$
add a comment |
$begingroup$
The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_3,3$ or $K_5$.
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
$endgroup$
add a comment |
$begingroup$
The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_3,3$ or $K_5$.
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
$endgroup$
The difficult part is not as difficult as some of the other answers, but perhaps the "if and only if" formulation is more natural: Kuratowski's theorem that a graph is planar if and only if it does not have a subgraph which is a subdivision of either $K_3,3$ or $K_5$.
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
edited Jan 11 at 15:18
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
answered Jan 11 at 9:33
Especially LimeEspecially Lime
21.9k22858
21.9k22858
add a comment |
add a comment |
$begingroup$
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
$endgroup$
2
$begingroup$
Assuming they're all non-empty.
$endgroup$
– Arnaud D.
Jan 11 at 9:14
$begingroup$
Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
$endgroup$
– Marc Paul
Jan 14 at 22:59
add a comment |
$begingroup$
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
$endgroup$
2
$begingroup$
Assuming they're all non-empty.
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– Arnaud D.
Jan 11 at 9:14
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Actually, the projections always being surjective is also equivalent to the axiom of choice.
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– Marc Paul
Jan 11 at 12:30
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@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
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– BigbearZzz
Jan 14 at 22:00
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@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
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– Marc Paul
Jan 14 at 22:59
add a comment |
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The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
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The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
answered Jan 11 at 6:45
Carlos JiménezCarlos Jiménez
2,3851520
2,3851520
2
$begingroup$
Assuming they're all non-empty.
$endgroup$
– Arnaud D.
Jan 11 at 9:14
$begingroup$
Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
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– Marc Paul
Jan 14 at 22:59
add a comment |
2
$begingroup$
Assuming they're all non-empty.
$endgroup$
– Arnaud D.
Jan 11 at 9:14
$begingroup$
Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
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– Marc Paul
Jan 14 at 22:59
2
2
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Assuming they're all non-empty.
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– Arnaud D.
Jan 11 at 9:14
$begingroup$
Assuming they're all non-empty.
$endgroup$
– Arnaud D.
Jan 11 at 9:14
$begingroup$
Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
Actually, the projections always being surjective is also equivalent to the axiom of choice.
$endgroup$
– Marc Paul
Jan 11 at 12:30
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@MarcPaul Would you mind explain a bit on why that is the case? I thought the projection of a product set onto its component looks pretty constructive to me.
$endgroup$
– BigbearZzz
Jan 14 at 22:00
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
$endgroup$
– Marc Paul
Jan 14 at 22:59
$begingroup$
@BgbearZzz Axiom of choice is equivalent to the statement that a product of non-empty sets is non-empty. So if AoC fails, the image of the projection map can be empty. And indeed, equipping each set with the discrete topology, and making sure that at least one of the sets is infinite, you obtain a contradiction to the 'easy' direction of the statement.
$endgroup$
– Marc Paul
Jan 14 at 22:59
add a comment |
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An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
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6
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There may be chances that Goldbach's conjecture is wrong...
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– YuiTo Cheng
Jan 11 at 9:20
add a comment |
$begingroup$
An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
$endgroup$
6
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
$endgroup$
– YuiTo Cheng
Jan 11 at 9:20
add a comment |
$begingroup$
An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
$endgroup$
An even integer $n$ is the sum of two primes iff $n>2$.
("If" is Goldbach's conjecture, which is still open. "Only if" is trivial.)
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
answered Jan 11 at 9:17
Rosie FRosie F
1,301315
1,301315
6
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
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– YuiTo Cheng
Jan 11 at 9:20
add a comment |
6
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
$endgroup$
– YuiTo Cheng
Jan 11 at 9:20
6
6
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
$endgroup$
– YuiTo Cheng
Jan 11 at 9:20
$begingroup$
There may be chances that Goldbach's conjecture is wrong...
$endgroup$
– YuiTo Cheng
Jan 11 at 9:20
add a comment |
$begingroup$
$mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
answered Jan 11 at 12:30
guest9366710guest9366710
111
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add a comment |
$begingroup$
$mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
answered Jan 11 at 12:30
guest9366710guest9366710
111
$endgroup$
add a comment |
$begingroup$
$mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
answered Jan 11 at 12:30
guest9366710guest9366710
111
$endgroup$
$mathbb R^n$ has the structure of a real division algebra iff $n=1,2,4$ or $8$.
answered Jan 11 at 12:30
guest9366710guest9366710
111
answered Jan 11 at 12:30
guest9366710guest9366710
111
answered Jan 11 at 12:30
guest9366710guest9366710
111
answered Jan 11 at 12:30
guest9366710guest9366710
111
111
add a comment |
add a comment |
3
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One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
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– Blue
Jan 11 at 5:21
2
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That's a good metaphor!
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– YuiTo Cheng
Jan 11 at 5:23
11
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The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
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– bof
Jan 11 at 5:39
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@bof done, see math.stackexchange.com/questions/3069590/…
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– YuiTo Cheng
Jan 11 at 7:57
3
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You cannot make such a question more specific. You are asking for a list of things of a certain nature, where that set is known to be enormous. Any answer is either just one among many with no way to determine a clear "best" or requires thousands of pages to sufficiently answer in one go. This is textbook classic VTC:TB.
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– Nij
Jan 11 at 9:57