If \(\mathbb{R}^\omega\) is given the product topology, show there is no continuous surjective map \(f: \mathbb{R} \to \mathbb{R}^\omega\). [Hint: Show that \(\mathbb{R}^\omega\) is not a countable union of compact subspaces.]
If \(\mathbb{R}^\omega\) is given the product topology, determine whether or not there is a continuous surjective map of \(\mathbb{R}\) onto the subspace \(\mathbb{R}^\infty\).
What happens to the statements in (a) and (b) if \(\mathbb{R}^\omega\) is given the uniform topology or the box topology?
Solution
- The real line is \(\sigma\)-compact: \[ \mathbb R=\bigcup_{n\ge1}[-n,n]. \] Hence every continuous image of \(\mathbb R\) is a countable union of compact subspaces.
But \(\mathbb R^\omega\) in the product topology is not \(\sigma\)-compact. Indeed, suppose \[ \mathbb R^\omega=\bigcup_{n\ge1}K_n \] with each \(K_n\) compact. The coordinate projection \(\pi_n(K_n)\subset\mathbb R\) is compact, hence bounded. Choose \(x_n\in\mathbb R-\pi_n(K_n)\), and let \(x=(x_1,x_2,\ldots)\). Then \(x\notin K_n\) for every \(n\), contradiction. Therefore no continuous surjection \(\mathbb R\to\mathbb R^\omega\) exists in the product topology.
There is a continuous surjection onto \(\mathbb R^\infty\). Let \[ K_m=[-m,m]^m\times\{0\}\times\{0\}\times\cdots. \] The \(K_m\)’s exhaust \(\mathbb R^\infty\). For each \(m\), choose a continuous surjection \(q_m:I\to K_m\) with \(q_m(0)=q_m(1)=0\), as in E-WVVWW. Map the interval \([m-1,m]\) onto \(K_m\) via \(q_m\). Since adjacent pieces agree at the origin, the pasted map \([0,\infty)\to\mathbb R^\infty\) is continuous and surjective; compose with \(t\mapsto|t|\) to obtain a continuous surjection from \(\mathbb R\).
In both the uniform and box topologies, \(\mathbb R^\omega\) is nonseparable. The subset \[ D=\{0,1\}^\omega \] is uncountable, and distinct points of \(D\) have uniform distance \(1\). Thus the uniform balls of radius \(1/3\) about points of \(D\) are pairwise disjoint. The box topology is finer than the uniform topology, so these balls are also box-open. A separable space cannot contain an uncountable family of pairwise disjoint nonempty open sets. Hence neither version of \(\mathbb R^\omega\) is separable. Since a continuous image of the separable space \(\mathbb R\) is separable, there is no continuous surjection from \(\mathbb R\) onto \(\mathbb R^\omega\) in either topology.
For \(\mathbb R^\infty\), however, the construction in (b) works in both the uniform and box subspace topologies. Each finite-stage inclusion \[ \mathbb R^m\times\{0\}^{\{i>m\}}\hookrightarrow\mathbb R^\infty \] is continuous for both topologies, and the piecewise construction is locally finite on the domain \([0,\infty)\). Thus continuous surjections \[ \mathbb R\twoheadrightarrow\mathbb R^\infty \] exist for the product, uniform, and box topologies.