Lebesgue number and continuity of function
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I was given a theorem in the section continuity of function in the book:
Theorem: Given a closed interval $[a,b]$ and an open cover $phi$ of $[a,b]$, $exists lambda(phi)$ such that $forall A subset [a,b]$ with $d(A):=sup{|x_i-x_j||x_i,x_j in A} < lambda$, then there is $I in phi$ such that $A subset I$. First How to prove it? I was thinking $lambda := inf{|a_i-b_j|}$ where $a_i,b_j$ are end points of some intervals in $phi$.
Also, how is this theorem relate to continuity of functions?
I was thinking defining $f(x)=x$ on $[a,b]$ and then for all $epsilon>0$, there is an open cover $I$ of $[a,b]$ such that $inf{|a_i-b_i||(a_i,b_i)in I}<epsilon$. Ok then there is $lambda$ such that $forall A subset [a,b]$ with $d(A) < lambda$, then there is $S in I$ such that $max(A)-min(A) < epsilon$.
real-analysis analysis
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
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add a comment |
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I was given a theorem in the section continuity of function in the book:
Theorem: Given a closed interval $[a,b]$ and an open cover $phi$ of $[a,b]$, $exists lambda(phi)$ such that $forall A subset [a,b]$ with $d(A):=sup{|x_i-x_j||x_i,x_j in A} < lambda$, then there is $I in phi$ such that $A subset I$. First How to prove it? I was thinking $lambda := inf{|a_i-b_j|}$ where $a_i,b_j$ are end points of some intervals in $phi$.
Also, how is this theorem relate to continuity of functions?
I was thinking defining $f(x)=x$ on $[a,b]$ and then for all $epsilon>0$, there is an open cover $I$ of $[a,b]$ such that $inf{|a_i-b_i||(a_i,b_i)in I}<epsilon$. Ok then there is $lambda$ such that $forall A subset [a,b]$ with $d(A) < lambda$, then there is $S in I$ such that $max(A)-min(A) < epsilon$.
real-analysis analysis
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
$endgroup$
1
$begingroup$
See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 8:27
add a comment |
$begingroup$
I was given a theorem in the section continuity of function in the book:
Theorem: Given a closed interval $[a,b]$ and an open cover $phi$ of $[a,b]$, $exists lambda(phi)$ such that $forall A subset [a,b]$ with $d(A):=sup{|x_i-x_j||x_i,x_j in A} < lambda$, then there is $I in phi$ such that $A subset I$. First How to prove it? I was thinking $lambda := inf{|a_i-b_j|}$ where $a_i,b_j$ are end points of some intervals in $phi$.
Also, how is this theorem relate to continuity of functions?
I was thinking defining $f(x)=x$ on $[a,b]$ and then for all $epsilon>0$, there is an open cover $I$ of $[a,b]$ such that $inf{|a_i-b_i||(a_i,b_i)in I}<epsilon$. Ok then there is $lambda$ such that $forall A subset [a,b]$ with $d(A) < lambda$, then there is $S in I$ such that $max(A)-min(A) < epsilon$.
real-analysis analysis
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
$endgroup$
I was given a theorem in the section continuity of function in the book:
Theorem: Given a closed interval $[a,b]$ and an open cover $phi$ of $[a,b]$, $exists lambda(phi)$ such that $forall A subset [a,b]$ with $d(A):=sup{|x_i-x_j||x_i,x_j in A} < lambda$, then there is $I in phi$ such that $A subset I$. First How to prove it? I was thinking $lambda := inf{|a_i-b_j|}$ where $a_i,b_j$ are end points of some intervals in $phi$.
Also, how is this theorem relate to continuity of functions?
I was thinking defining $f(x)=x$ on $[a,b]$ and then for all $epsilon>0$, there is an open cover $I$ of $[a,b]$ such that $inf{|a_i-b_i||(a_i,b_i)in I}<epsilon$. Ok then there is $lambda$ such that $forall A subset [a,b]$ with $d(A) < lambda$, then there is $S in I$ such that $max(A)-min(A) < epsilon$.
real-analysis analysis
real-analysis analysis
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
asked Dec 17 '18 at 8:15
nafhgoodnafhgood
1,803422
1,803422
1
$begingroup$
See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 8:27
add a comment |
1
$begingroup$
See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 8:27
1
1
$begingroup$
See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 8:27
$begingroup$
See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 8:27
add a comment |
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See answer by Lierre to the following: math.stackexchange.com/questions/105337/…
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– Kavi Rama Murthy
Dec 17 '18 at 8:27