Exploring the Boundary of the 3-Sphere: A Topological Analysis

[jk22]

Btw : checking wikipedia about boundary :
https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took $S=[0,1]^2\cap\mathbb{Q}^2$

Is the following correct :

$\partial S=[0,1]^2$
$\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])$
$\partial\partial\partial S=\emptyset$ ?

 

[ChinleShale]

Btw : checking wikipedia about boundary :
https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took $S=[0,1]^2\cap\mathbb{Q}^2$

Is the following correct :

$\partial S=[0,1]^2$
$\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])$
$\partial\partial\partial S=\emptyset$ ?

How do you justify each these three steps?

 

[mathwonk]

"But what about the other way around? Suppose p is mapped to the interior of the upper half space by a local homeomorphism of an open set U. How does your proof show that U can not contain an open subset containing p that can be mapped homeomorpically to an open subset of the boundary of R+n and take p to the boundary?"

By shrinking around the boundary point, and composing, it seems this would give a homeomorphism from a half disc H around the boundary point y, to a bounded connected open set W in the interior, taking y to some point x in W. Then the complement of y in the half disc, i.e. the punctured half disc H - {y}, would be homeomorphic to the punctured set W-{x}. But one of them is contractible and the other, namely W - {x}, is not. I.e. a small sphere centered at x wraps once around x, and hence cannot be deformed off it, as a contraction would do.

I.e. as stated originally, y has arbitrarily small contractible punctured neighborhoods in H, while x has no contractible punctured nbhds in W. So H cannot be homeomorphic to W, with y going to x.
Perhaps I should say (H,y) and (W,x) have different local homology groups, i.e. the local homology of H at y is zero, unlike that of W at x.

believe that? sorry, I am a bit rusty on topology. But I am using excision here to compute the local homology of W at x, by replacing W by a ball centered at x. So even though originally I am trying to compare a punctured half ball to a punctured open set in R^n, excision let's me just compare a punctured half ball to a punctured ball.

 

[trees and plants]

Could the theorem for boundary of A, $\partial{A}=\overline{A}-{A}^\circ$ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

 

[fresh_42]

Could the theorem for boundary of A, $\partial{A}=\overline{A}-{A}^\circ$ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

No, because ...

The term boundary of a manifold has a different meaning than the boundary in a topological space.

You gave the definition of the latter whereas I assume the former is meant.

 

[ChinleShale]

Could the theorem for boundary of A, ∂A=A¯−A∘ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

As @fresh_42 says in post #42 the two definitions of boundary, the topological definition and the definition of the boundary of a manifold are not the same.

Here is an example: Start with the closed unit ball in $R^3$. This is a three dimensional manifold with boundary equal to the unit sphere.

As a subset of $R^3$ its interior is the open unit ball so its closure minus its interior is the unit sphere. This is the same as its boundary as a manifold rather than as a subset of $R^3$.

As a subset of itself though, it is both open and closed so its boundary is empty.

Suppose $R^3$ is thought of as the points in $R^4$ whose fourth coordinate is zero. Then the unit ball in $R^3$ becomes a subset of $R^4$. As a subset of $R^4$ it is closed and has empty interior. So its boundary is the entire unit ball.

 

[ChinleShale]

@infinitely small

Continuing with post #41

Notice that the only case in post #41 where the boundary of the solid ball was both its topological and manifold boundary was when it was a subset of 3 dimensional space. This is because open sets in the manifold 3 ball are also open sets in $R^3$. One might wonder whether this would work for an arbitrary manifold with boundary. Sadly, this is not possible in general. Even a three dimensional manifold with boundary may not be realizable as a subset of $R^3$. For instance, a 3 dimensional manifold whose boundary is a Klein bottle can not be a subset of $R^3$. More generally an n-dimensional manifold with boundary may only be realizable as a subset of a higher dimensional Euclidean space.

 

[trees and plants]

The boundary of a manifold is the complement of its interior and its interior is the set of all points in the manifold which have neighborhoods homeomorphic to an open subset of $R^n$ read if you want this https://en.wikipedia.org/wiki/Manifold where it says about manifold with boundary, boundary and interior.

Perhaps this helps? If we show that the complement of the interior of the 3-sphere is the empty set then the work is done.

 

[WWGD]

Cant we just use some homological/topological argument?

The 3-sphere is the boundary of the 3 -ball. The boundary of a boundary is empty. EDIT: We can use, e.g., Simplicial homology, then it follows dod =d^2 =0.(empty). Don't know if this is too over the top , i.e., unnecessarily complicated.