What are the components and path components of \(\mathbb{R}^\omega\) (in the product topology)?
Consider \(\mathbb{R}^\omega\) in the uniform topology. Show that \(\mathbf{x}\) and \(\mathbf{y}\) lie in the same component of \(\mathbb{R}^\omega\) if and only if the sequence
\[ \mathbf{x} - \mathbf{y} = (x_1 - y_1, x_2 - y_2, \dots) \]
is bounded. [Hint: It suffices to consider the case where \(\mathbf{y} = 0\).]
- Give \(\mathbb{R}^\omega\) the box topology. Show that \(\mathbf{x}\) and \(\mathbf{y}\) lie in the same component of \(\mathbb{R}^\omega\) if and only if the sequence \(\mathbf{x} - \mathbf{y}\) is “eventually zero.” [Hint: If \(\mathbf{x} - \mathbf{y}\) is not eventually zero, show there is a homeomorphism \(h\) of \(\mathbb{R}^\omega\) with itself such that \(h(\mathbf{x})\) is bounded and \(h(\mathbf{y})\) is unbounded.]
Solution
In the product topology, \(\mathbb R^\omega\) is path connected. Given \(x,y\in\mathbb R^\omega\), the straight-line path \[ \gamma(t)=((1-t)x_i+ty_i)_{i\ge1} \] is continuous coordinatewise, hence continuous into the product. Therefore there is one path component and one component: all of \(\mathbb R^\omega\).
In the uniform topology, translation is a homeomorphism, so it suffices to determine the component of \(0\). Suppose \(x=(x_i)\) is bounded, say \(|x_i|\le M\). The straight-line path \[ \gamma(t)=tx \] is uniformly continuous in the parameter with respect to the uniform metric, since for \(s,t\in[0,1]\), \[ \bar\rho(sx,tx)\le \min\{M|s-t|,1\}. \] Hence every bounded sequence lies in the path component of \(0\).
Conversely, let \[ B=\{x\in\mathbb R^\omega:x\text{ is bounded}\}. \] For any \(x\in B\), every sufficiently small uniform ball about \(x\) consists of bounded sequences; for any unbounded \(x\), every uniform ball of radius \(<1\) consists of unbounded sequences, since a uniformly bounded perturbation cannot turn an unbounded sequence into a bounded one. Thus \(B\) is both open and closed in the uniform topology. The component of \(0\) is therefore contained in \(B\). Hence two points \(x,y\) lie in the same component exactly when \[ x-y \] is bounded; in fact these components are path components as well.
- In the box topology, if \(x-y\) is eventually zero, then the two points differ in only finitely many coordinates. The straight-line path between them varies only in these finitely many coordinates, so it is continuous in the box topology. Thus they lie in the same path component.
Conversely, suppose \(d=x-y\) is not eventually zero. By translation, take \(y=0\). Choose infinitely many indices \(i_k\) with \(d_{i_k}\ne0\). Define positive scalars \(a_i\) by \[ a_{i_k}=\frac{k}{|d_{i_k}|}, \] and \(a_i=1\) at all other indices. The coordinatewise scaling \[ h((z_i))=(a_i z_i) \] is a homeomorphism of the box product (its inverse scales by \(a_i^{-1}\)). Then \[ h(0)=0 \] is bounded, while \(h(d)\) is unbounded since its \(i_k\)th coordinate has absolute value \(k\).
If \(0\) and \(d\) lay in one box-connected subset, their images under \(h\) would lie in one box-connected subset. Since the box topology is finer than the uniform topology, every box-connected set is also connected in the uniform topology. But part (b) says a bounded point and an unbounded point lie in different uniform components. Contradiction. Hence \(x,y\) are in the same box component iff \(x-y\) is eventually zero; these are again exactly the path components.