Boundary of $\mathbb{R}^4$ and fundamental group of $\mathbb{R}^4/\mathbb{R}^2$

Question

a) To what I know boundary makes no sense in open sets. Does it make any sense to talk about the boundary of $\mathbb{R}^4$? In physics they consider it when discussing wether the universe has a boundary.

b) How can I calculate the fundamental group of $\mathbb{R}^4/\mathbb{R}^2$? I know the rule for product spaces but I have no idea for this case!

Answer 1

There's another definition of boundary for maniforld, which are spaces where every point has a neighborhood homeomorphic to $\mathbb R^n$ or $\mathbb R_+^n$, for a fixed $n$. The boundary of a manifold is the set of points with neighborhood homeomorphic to $\mathbb R_+^n$, and indeed for $\mathbb R^n$, there is no boundary.

Answer 2

Usually, topologists talk about boundaries of a subset of another set. So, if your $\mathbb{R}^4$ sits in a larger space -- for example, in $S^4$ -- it may have a boundary (one point, for example).

There is another definition of boundary for manifolds: in this sense, $\mathbb{R}^4$ has no boundary.

As far as I know, physicists have various theories and assumptions about the topology of the universe and I believe most of them are boundaryless, although they are not always as simple as $\mathbb{R}^4$ and not even 4-dimensional.

As for the second question, the fundamental group of $\mathbb{R}^4/\mathbb{R^2}$ is trivial, because you can homotop a curve to the point $\mathbb{R^2}$.

EDIT: To expand the last step, note that $\mathbb{R}^4/\mathbb{R}^2$ is contractible. Let $\mathbb{R}^2$ be the linear subspace $\{(*,*,0,0)\}$ and $\pi$ be the quotient map $\pi: \mathbb{R}^4\to\mathbb{R}^4/\mathbb{R}^2$. Let $F: \mathbb{R}^4\times[0,1]\to \mathbb{R}^4$ be defined by $F(x_1, x_2, x_3, x_4, t)=(x_1, x_2, t\,x_3, t\,x_4)$. This a a deformation retraction (time $t$ goes from $1$ to $0$) of $\mathbb{R}^4$ to $\mathbb{R}^2$ fixing $\mathbb{R}^2$, so the composition $\pi\circ F$ is a (strong) deformation retraction of $\mathbb{R}^4/\mathbb{R}^2$ to the point $\mathbb{R}^2 \in\mathbb{R}^4/\mathbb{R}^2$.

Answer 3

Every set has a (possibly empty) boundary, whether or not it is open.

Let $X$ be a topological space, say $\Bbb R^4$ with standard topology. If $A$ is an arbitrary subset of $X$, then the boundary of $A$ is the set of points that are "near" to both $A$ and $X \setminus A$ and is sometimes defined by $\partial A = \operatorname{cl}(A) \setminus \operatorname{int}(A)$.

Using this, one can see that $\partial\Bbb R^4 = \emptyset$, and in general, $\partial X = \emptyset$. So in some sense these sets have no boundary, but to be more precise, they have no boundary points.

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Be aware that the boundary of a set depends on what topology you are considering it under. For example, if $(0,1) \subseteq \Bbb R$ in the usual topology, then it has boundary $\{0,1\}$, however, in the subspace topology, $(0,1)$ has empty boundary.