Definition of Shrinks Nicely
The following is the definition of shrinks nicely according to Folland:
a family ${E_r}_{r>0}$ of subsets of $mathbb{R}^n$ shrinks nicely to $x$ if $E_r subset B(x,r)$ for all $r>0$ and $m(E_r) > alpha m(B(x,r))$ for all $r>0$, for some universal constant $alpha$.
I am confused about the indexing of the a family ${E_r}$. Can there be a single E1 and that is the only set in the family? Must the r converge to 0? Must the index be uncountable?
Thank you for the help
real-analysis
asked Nov 25 at 0:04
user56628
578
add a comment |
The following is the definition of shrinks nicely according to Folland:
a family ${E_r}_{r>0}$ of subsets of $mathbb{R}^n$ shrinks nicely to $x$ if $E_r subset B(x,r)$ for all $r>0$ and $m(E_r) > alpha m(B(x,r))$ for all $r>0$, for some universal constant $alpha$.
I am confused about the indexing of the a family ${E_r}$. Can there be a single E1 and that is the only set in the family? Must the r converge to 0? Must the index be uncountable?
Thank you for the help
real-analysis
asked Nov 25 at 0:04
user56628
578
1
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55
add a comment |
The following is the definition of shrinks nicely according to Folland:
a family ${E_r}_{r>0}$ of subsets of $mathbb{R}^n$ shrinks nicely to $x$ if $E_r subset B(x,r)$ for all $r>0$ and $m(E_r) > alpha m(B(x,r))$ for all $r>0$, for some universal constant $alpha$.
I am confused about the indexing of the a family ${E_r}$. Can there be a single E1 and that is the only set in the family? Must the r converge to 0? Must the index be uncountable?
Thank you for the help
real-analysis
asked Nov 25 at 0:04
user56628
578
The following is the definition of shrinks nicely according to Folland:
a family ${E_r}_{r>0}$ of subsets of $mathbb{R}^n$ shrinks nicely to $x$ if $E_r subset B(x,r)$ for all $r>0$ and $m(E_r) > alpha m(B(x,r))$ for all $r>0$, for some universal constant $alpha$.
I am confused about the indexing of the a family ${E_r}$. Can there be a single E1 and that is the only set in the family? Must the r converge to 0? Must the index be uncountable?
Thank you for the help
real-analysis
real-analysis
asked Nov 25 at 0:04
user56628
578
asked Nov 25 at 0:04
user56628
578
asked Nov 25 at 0:04
user56628
578
asked Nov 25 at 0:04
user56628
578
asked Nov 25 at 0:04
user56628
578
578
1
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55
add a comment |
1
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55
1
1
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55
add a comment |
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1
As stated, the family must be indexed by $(0, infty)$, so there is a set $E_r$ for each $r > 0$. I am not sure what the context is, but I think one can relax this definition by considering families indexed by any set of positive reals $r$ that has zero as a limit point.
– angryavian
Nov 25 at 0:09
(a "just passing by" comment) I wonder why Folland uses the potentially misleading phrase "for some universal constant $alpha$?" When I see this phrase, my thoughts immediately go to something like $pi,$ or $e,$ or $ln (pi - sqrt{5}),$ etc. and only after spending a few moments rereading the definition and thinking of what makes sense do I realize all that is intended is that the existential quantifier "there exists $alpha$" comes before the universal quantifier "for all $r,$ and then I waste time wondering why the author didn't just phrase it that way. (continued)
– Dave L. Renfro
Nov 25 at 8:49
For example, why not just say "there exists $alpha > 0$ such that for each $r>0$ we have $E_r subset B(x,r)$ and $m(E_r) > alpha m(B(x,r))$"? This is shorter and easier to parse mathematically, even with the additional assumption that $alpha$ is positive (which I assume Folland intended, although maybe I'm wrong).
– Dave L. Renfro
Nov 25 at 8:55