Proof of Rudin's Theorem 3.10If $p$ is an element of $overline E$ but not a limit point of $E$, then why is...
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Proof of Rudin's Theorem 3.10
If $p$ is an element of $overline E$ but not a limit point of $E$, then why is there a $p' in E$ such that $d(p, p') < varepsilon$?The diameter of a compact set.Use of epsilon in proofs about diametersProve that $exists a in X$ s.t. $bigcap ^{infty} _{n=1} F_n = {a}$Which metric is used in this limitLet (X,d) is metric space. Is it correct diam(A)=diam(co(A))Probs. 21 and 22 in Baby Rudin: Baire's TheoremCauchy sequences and diameter of a setProof of Rudin's Theorem 3.11 (c)Any Isodiametric-type inequality for diameter and perimeter in high dimension?
$begingroup$
Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,qin E$. The sup of $S$ is called the diameter of $E$.
Theorem 3.10. If $overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$text{diam }overline{E} = text{diam }E.$$
Proof: Fix $varepsilon>0$, and choose $p, q in overline{E}$. By the definition of $overline{E}$, there are points $p',q' in E$ such that $d(p,p') < varepsilon$ and $d(q,q') < varepsilon$. Hence $$d(p, q) le d(p,p') + d(p', q') + d(q', q) < 2varepsilon + d(p', q') le 2varepsilon + text{diam }E.$$
Ok until here. But then they use the inequality above to come up with $$text{diam }overline{E} le 2varepsilon + text{diam }E$$
Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.
real-analysis inequality proof-explanation
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
$endgroup$
add a comment |
$begingroup$
Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,qin E$. The sup of $S$ is called the diameter of $E$.
Theorem 3.10. If $overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$text{diam }overline{E} = text{diam }E.$$
Proof: Fix $varepsilon>0$, and choose $p, q in overline{E}$. By the definition of $overline{E}$, there are points $p',q' in E$ such that $d(p,p') < varepsilon$ and $d(q,q') < varepsilon$. Hence $$d(p, q) le d(p,p') + d(p', q') + d(q', q) < 2varepsilon + d(p', q') le 2varepsilon + text{diam }E.$$
Ok until here. But then they use the inequality above to come up with $$text{diam }overline{E} le 2varepsilon + text{diam }E$$
Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.
real-analysis inequality proof-explanation
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
$endgroup$
$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
1
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
1
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27
add a comment |
$begingroup$
Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,qin E$. The sup of $S$ is called the diameter of $E$.
Theorem 3.10. If $overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$text{diam }overline{E} = text{diam }E.$$
Proof: Fix $varepsilon>0$, and choose $p, q in overline{E}$. By the definition of $overline{E}$, there are points $p',q' in E$ such that $d(p,p') < varepsilon$ and $d(q,q') < varepsilon$. Hence $$d(p, q) le d(p,p') + d(p', q') + d(q', q) < 2varepsilon + d(p', q') le 2varepsilon + text{diam }E.$$
Ok until here. But then they use the inequality above to come up with $$text{diam }overline{E} le 2varepsilon + text{diam }E$$
Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.
real-analysis inequality proof-explanation
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
$endgroup$
Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,qin E$. The sup of $S$ is called the diameter of $E$.
Theorem 3.10. If $overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$text{diam }overline{E} = text{diam }E.$$
Proof: Fix $varepsilon>0$, and choose $p, q in overline{E}$. By the definition of $overline{E}$, there are points $p',q' in E$ such that $d(p,p') < varepsilon$ and $d(q,q') < varepsilon$. Hence $$d(p, q) le d(p,p') + d(p', q') + d(q', q) < 2varepsilon + d(p', q') le 2varepsilon + text{diam }E.$$
Ok until here. But then they use the inequality above to come up with $$text{diam }overline{E} le 2varepsilon + text{diam }E$$
Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.
real-analysis inequality proof-explanation
real-analysis inequality proof-explanation
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
asked May 23 '17 at 0:01
AnalyticHarmonyAnalyticHarmony
672313
672313
$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
1
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
1
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27
add a comment |
$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
1
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
1
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27
$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
1
1
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
1
1
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're missing an important point. If $x<c$ for all $xin S$, then $sup Sle c$. (For example, take $S = (0,1)$. For every $xin S$, we have $x<1$, but $sup S = 1$.)
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
add a comment |
$begingroup$
Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q in E$ such that $d(p, p') < epsilon$, $d(q, q') < epsilon$ is the following:
Every point in $overline{E}$ is either
(1.) a limit point of $E$
(2.) a point in $E$
If $p$ is a limit point of $E$, then we can find a point $p' in N_{epsilon}(p)$ such that $p' not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $epsilon$, $d(p, p') < epsilon$.
If $p in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < epsilon$.
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
You're missing an important point. If $x<c$ for all $xin S$, then $sup Sle c$. (For example, take $S = (0,1)$. For every $xin S$, we have $x<1$, but $sup S = 1$.)
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
add a comment |
$begingroup$
You're missing an important point. If $x<c$ for all $xin S$, then $sup Sle c$. (For example, take $S = (0,1)$. For every $xin S$, we have $x<1$, but $sup S = 1$.)
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
add a comment |
$begingroup$
You're missing an important point. If $x<c$ for all $xin S$, then $sup Sle c$. (For example, take $S = (0,1)$. For every $xin S$, we have $x<1$, but $sup S = 1$.)
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
You're missing an important point. If $x<c$ for all $xin S$, then $sup Sle c$. (For example, take $S = (0,1)$. For every $xin S$, we have $x<1$, but $sup S = 1$.)
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
answered May 23 '17 at 0:12
Ted ShifrinTed Shifrin
64.3k44692
64.3k44692
add a comment |
add a comment |
$begingroup$
Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q in E$ such that $d(p, p') < epsilon$, $d(q, q') < epsilon$ is the following:
Every point in $overline{E}$ is either
(1.) a limit point of $E$
(2.) a point in $E$
If $p$ is a limit point of $E$, then we can find a point $p' in N_{epsilon}(p)$ such that $p' not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $epsilon$, $d(p, p') < epsilon$.
If $p in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < epsilon$.
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
$endgroup$
add a comment |
$begingroup$
Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q in E$ such that $d(p, p') < epsilon$, $d(q, q') < epsilon$ is the following:
Every point in $overline{E}$ is either
(1.) a limit point of $E$
(2.) a point in $E$
If $p$ is a limit point of $E$, then we can find a point $p' in N_{epsilon}(p)$ such that $p' not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $epsilon$, $d(p, p') < epsilon$.
If $p in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < epsilon$.
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
$endgroup$
add a comment |
$begingroup$
Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q in E$ such that $d(p, p') < epsilon$, $d(q, q') < epsilon$ is the following:
Every point in $overline{E}$ is either
(1.) a limit point of $E$
(2.) a point in $E$
If $p$ is a limit point of $E$, then we can find a point $p' in N_{epsilon}(p)$ such that $p' not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $epsilon$, $d(p, p') < epsilon$.
If $p in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < epsilon$.
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
$endgroup$
Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q in E$ such that $d(p, p') < epsilon$, $d(q, q') < epsilon$ is the following:
Every point in $overline{E}$ is either
(1.) a limit point of $E$
(2.) a point in $E$
If $p$ is a limit point of $E$, then we can find a point $p' in N_{epsilon}(p)$ such that $p' not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $epsilon$, $d(p, p') < epsilon$.
If $p in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < epsilon$.
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
edited Mar 10 at 13:23
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
answered Mar 10 at 13:13
James ShapiroJames Shapiro
1938
1938
add a comment |
add a comment |
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$begingroup$
Definition of closure of $E$: $overline{E} = E cup E' $, where $E'$ is the set of limit points of $E$.
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:03
1
$begingroup$
Doesn't $<$ imply $leq$?
$endgroup$
– avs
May 23 '17 at 0:03
$begingroup$
Yes, but I think if you have $A ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition?
$endgroup$
– AnalyticHarmony
May 23 '17 at 0:05
1
$begingroup$
Ok, you seem comfortable getting to this: for every $epsilon > 0$ and for all $p, q in overline{E}$, one has $$ d(p, q) < text{diam }E + 2 epsilon $$ Doesn't that imply that $$ sup_{p, q in overline{E}} d(p, q) leq text{diam }E + 2 epsilon? $$ The strict inequality becomes a sharp one upon taking the $sup$ on the left-hand side.
$endgroup$
– avs
May 23 '17 at 0:27