Skip to main content
Bumped by Community user
added 234 characters in body
Source Link
Clem.
  • 353
  • 1
  • 10

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

You can write $\overline{B}(x,r)$ as the non-increasing intersection of the open balls $B(x,s)$ for $s>r$, which are homeomorphic to an open disc; maybe this could help...

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

An idea might be to find a characterization of a set homeomorphic to a closed disc. For example with a property of its interior, or its boundary, or its homotopy groups, or something else... I would be very interested by any idea!

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

You can write $\overline{B}(x,r)$ as the non-increasing intersection of the open balls $B(x,s)$ for $s>r$, which are homeomorphic to an open disc; maybe this could help...

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

You can write $\overline{B}(x,r)$ as the non-increasing intersection of the open balls $B(x,s)$ for $s>r$, which are homeomorphic to an open disc; maybe this could help...

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

An idea might be to find a characterization of a set homeomorphic to a closed disc. For example with a property of its interior, or its boundary, or its homotopy groups, or something else... I would be very interested by any idea!

added 175 characters in body
Source Link
Clem.
  • 353
  • 1
  • 10

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

You can write $\overline{B}(x,r)$ as the non-increasing intersection of the open balls $B(x,s)$ for $s>r$, which are homeomorphic to an open disc; maybe this could help...

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

Let $\Sigma$ be a compact (smooth) surface, with a geodesic metric $d$ (compatible with the topology of $\Sigma$).

Let $x \in \Sigma$, and suppose you have the following: for every $r<1$, the open balls $B(x,r)=\{y \in \Sigma, d(x,y)<r\}$ are homeomorphic to an open disc on the plane.

Then, is the following true: for every $r<1$, the closed balls $\overline{B}(x,r)=\{y \in \Sigma,d(x,y) \leq r\}$ are homeomorphic to a closed disk on the plane ? (since the metric is intrisic, $\overline{B}(x,r)$ is the closure of $B(x,r)$).

I think it is true, but I can not find any proof. We can see this as a problem about the topology of closed sets in the plane, but I don't know much about it.

You can write $\overline{B}(x,r)$ as the non-increasing intersection of the open balls $B(x,s)$ for $s>r$, which are homeomorphic to an open disc; maybe this could help...

PS : you can find easy examples showing that neither "$B(x,r)$ is homeomorphic to an open disc" nor "$\overline{B}(x,r)$ is homeomorphic to a closed disc" is a consequence of the other.

edited title
Link
Clem.
  • 353
  • 1
  • 10

Open balls (resp., closed balls) balls homeomorphic to open (resp., closed) ballsdiscs on the plane

Source Link
Clem.
  • 353
  • 1
  • 10
Loading