Skip to main content
added 64 characters in body
Source Link

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

(finite perimeter set is on the distributional sense of GMT)

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

(finite perimeter set is on the distributional sense of GMT)

added 19 characters in body
Source Link

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

Are smooth compact surfaces embedded in R3 , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

Are smooth compact surfaces embedded in R3 (with no boundary) , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

added 84 characters in body
Source Link

Are smooth compact surfaces embedded in R3 , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections; I don't know if there isintersections. There are some problemproblems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

Are smooth surfaces embedded in R3 , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections; I don't know if there is some problem with the orientability.

Are smooth compact surfaces embedded in R3 , with bounded area, always the boundary of a finite perimeter set?

Take a smooth surface S in R3 embedded , with finite area. Can we say that the "interior" A of S is a Borel set? If yes, one can use the Gauss-Green Theorem to say that A has finite perimeter.

For example the sphere S^2 in R3 can be seen as the boundary of the open ball in R3; and so one can think of the ball as a finite perimeter set.

The bound on the area is fundamental, the embedded property assure that one can't have some issues with intersections. There are some problems with the orientability: for example the Mobius strip can be a counterexample if one allows S to be non-compact.

Source Link
Loading