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Why is my textbook using a rather long proof involving sequences?
Clash Royale CLAN TAG#URR8PPP
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Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $exists k_iin K_i$ such that $0<vert k_1-k_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$.
Can't we simply say that if $0=vert y_1-y_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$, then $y_1in K_1cap K_2$ - a contradiction?
here what the book provided:
real-analysis proof-writing
asked Jul 22 at 5:45
TheLast Cipher
538414
 |Â
show 1 more comment
up vote
2
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favorite
Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $exists k_iin K_i$ such that $0<vert k_1-k_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$.
Can't we simply say that if $0=vert y_1-y_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$, then $y_1in K_1cap K_2$ - a contradiction?
here what the book provided:
real-analysis proof-writing
asked Jul 22 at 5:45
TheLast Cipher
538414
2
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
2
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
4
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
1
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
2
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $exists k_iin K_i$ such that $0<vert k_1-k_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$.
Can't we simply say that if $0=vert y_1-y_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$, then $y_1in K_1cap K_2$ - a contradiction?
here what the book provided:
real-analysis proof-writing
asked Jul 22 at 5:45
TheLast Cipher
538414
Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $exists k_iin K_i$ such that $0<vert k_1-k_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$.
Can't we simply say that if $0=vert y_1-y_2vert=infvert x_1-x_2vert : x_1in K_1 quadlandquad x_2in K_2$, then $y_1in K_1cap K_2$ - a contradiction?
here what the book provided:
real-analysis proof-writing
asked Jul 22 at 5:45
TheLast Cipher
538414
asked Jul 22 at 5:45
TheLast Cipher
538414
asked Jul 22 at 5:45
TheLast Cipher
538414
asked Jul 22 at 5:45
TheLast Cipher
538414
538414
2
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
2
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
4
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
1
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
2
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58
 |Â
show 1 more comment
2
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
2
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
4
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
1
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
2
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58
2
2
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
2
2
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
4
4
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
1
1
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
2
2
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58
 |Â
show 1 more comment
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A ge B = C$ (a shorthand for $A ge B$ and $B = C$)
But that is not the case : its negation is $A ge B$ or $B neq C$.
So it is not valid to prove $A < B = C$ by assuming $A ge B = C$ and getting a contradiction.
answered Jul 22 at 7:35
mercio
43.9k256107
add a comment |Â
up vote
4
down vote
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 times K_2$ is compact by Tychonoff. The function $K_1 times K_2 to Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 in K_1$ and $k_2 in K_2$.
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A ge B = C$ (a shorthand for $A ge B$ and $B = C$)
But that is not the case : its negation is $A ge B$ or $B neq C$.
So it is not valid to prove $A < B = C$ by assuming $A ge B = C$ and getting a contradiction.
answered Jul 22 at 7:35
mercio
43.9k256107
add a comment |Â
up vote
3
down vote
accepted
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A ge B = C$ (a shorthand for $A ge B$ and $B = C$)
But that is not the case : its negation is $A ge B$ or $B neq C$.
So it is not valid to prove $A < B = C$ by assuming $A ge B = C$ and getting a contradiction.
answered Jul 22 at 7:35
mercio
43.9k256107
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A ge B = C$ (a shorthand for $A ge B$ and $B = C$)
But that is not the case : its negation is $A ge B$ or $B neq C$.
So it is not valid to prove $A < B = C$ by assuming $A ge B = C$ and getting a contradiction.
answered Jul 22 at 7:35
mercio
43.9k256107
$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A ge B = C$ (a shorthand for $A ge B$ and $B = C$)
But that is not the case : its negation is $A ge B$ or $B neq C$.
So it is not valid to prove $A < B = C$ by assuming $A ge B = C$ and getting a contradiction.
answered Jul 22 at 7:35
mercio
43.9k256107
answered Jul 22 at 7:35
mercio
43.9k256107
answered Jul 22 at 7:35
mercio
43.9k256107
answered Jul 22 at 7:35
mercio
43.9k256107
43.9k256107
add a comment |Â
add a comment |Â
up vote
4
down vote
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 times K_2$ is compact by Tychonoff. The function $K_1 times K_2 to Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 in K_1$ and $k_2 in K_2$.
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
add a comment |Â
up vote
4
down vote
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 times K_2$ is compact by Tychonoff. The function $K_1 times K_2 to Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 in K_1$ and $k_2 in K_2$.
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 times K_2$ is compact by Tychonoff. The function $K_1 times K_2 to Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 in K_1$ and $k_2 in K_2$.
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
Here's a short proof if you want it:
$K_1$ and $K_2$ are compact, so $K_1 times K_2$ is compact by Tychonoff. The function $K_1 times K_2 to Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 in K_1$ and $k_2 in K_2$.
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
answered Jul 22 at 6:37
Kenny Lau
18.7k2157
18.7k2157
add a comment |Â
add a comment |Â
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2
But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1in K_1$ and $y_2in K_2$. However, this is required by the claim.
– Suzet
Jul 22 at 5:50
2
No. You need to use compactness somehow. e.g., Consider $K_1 = 0$ and $K_2 = 1/n $.
– Jair Taylor
Jul 22 at 5:50
4
The non-trivial claim is that the $inf$ is attained by some $k_1, k_2$, not that the $inf$ is strictly positive
– Rhys Steele
Jul 22 at 5:50
1
I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $Bbb R$ is closed.
– Lord Shark the Unknown
Jul 22 at 5:56
2
inf is not min. You have no idea that any $|y_1 - y_2| = inf$ exist.
– fleablood
Jul 22 at 5:58