Show that For $A$ closed in a metric space $(π,π)$ and $a$ isn't in A, $π(π₯,A):=inf[π(a,x) : x in A]$ > 0Prove [$A subset B$, and $B$ is bounded in $n$-space] $implies$ [ diameter of $A leq$ diameter of $B$]distance between a point and a set and open/closed setsProve the distance function in a metric space is continousIs the following function uniformly continuous?Prove that if $A cap B = emptyset$ and if dist($A,B$) = inf $rho (x,y) : x in A$ and $ y in B$ then dist($A,B) > 0$.Trying to follow proof in Royden that every metric space is normalDistance between two sets, understanding least upper bound property and applications of itShow that infimum of set is in closure of set.Metric Spaces: Explore the relationship between $partial_P S$ and $partial_X_i pi_i(S)$ for $i in mathbb N_n.$dist$(x,A)=0$ if and only if $xin overlineA$ (closure of $A$)
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Show that For $A$ closed in a metric space $(𝑋,𝑑)$ and $a$ isn't in A, $𝑑(𝑥,A):=inf[𝑑(a,x) : x in A]$ > 0
Prove [$A subset B$, and $B$ is bounded in $n$-space] $implies$ [ diameter of $A leq$ diameter of $B$]distance between a point and a set and open/closed setsProve the distance function in a metric space is continousIs the following function uniformly continuous?Prove that if $A cap B = emptyset$ and if dist($A,B$) = inf $rho (x,y) : x in A$ and $ y in B$ then dist($A,B) > 0$.Trying to follow proof in Royden that every metric space is normalDistance between two sets, understanding least upper bound property and applications of itShow that infimum of set is in closure of set.Metric Spaces: Explore the relationship between $partial_P S$ and $partial_X_i pi_i(S)$ for $i in mathbb N_n.$dist$(x,A)=0$ if and only if $xin overlineA$ (closure of $A$)
$begingroup$
I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.
Can I just say:
Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$
real-analysis analysis proof-verification
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
$endgroup$
|
show 3 more comments
$begingroup$
I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.
Can I just say:
Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$
real-analysis analysis proof-verification
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
$endgroup$
$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55
|
show 3 more comments
$begingroup$
I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.
Can I just say:
Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$
real-analysis analysis proof-verification
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
$endgroup$
I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.
Can I just say:
Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$
real-analysis analysis proof-verification
real-analysis analysis proof-verification
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
edited Mar 13 at 13:20
David C. Ullrich
61.5k43994
61.5k43994
asked Mar 13 at 10:44
70pr4k70pr4k
678
asked Mar 13 at 10:44
70pr4k70pr4k
678
asked Mar 13 at 10:44
70pr4k70pr4k
678
678
$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55
|
show 3 more comments
$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55
$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.
answered Mar 13 at 11:01
BerciBerci
61.5k23674
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.
answered Mar 13 at 11:01
BerciBerci
61.5k23674
$endgroup$
add a comment |
$begingroup$
Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.
answered Mar 13 at 11:01
BerciBerci
61.5k23674
$endgroup$
add a comment |
$begingroup$
Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.
answered Mar 13 at 11:01
BerciBerci
61.5k23674
$endgroup$
Assume $d(A, a) =0$.
By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.
But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.
answered Mar 13 at 11:01
BerciBerci
61.5k23674
answered Mar 13 at 11:01
BerciBerci
61.5k23674
answered Mar 13 at 11:01
BerciBerci
61.5k23674
answered Mar 13 at 11:01
BerciBerci
61.5k23674
61.5k23674
add a comment |
add a comment |
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$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51
$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51
$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52
$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53
$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55