Consider the product, uniform, and box topologies on \(\mathbb{R}^\omega\).
- In which topologies are the following functions from \(\mathbb{R}\) to \(\mathbb{R}^\omega\) continuous?
\[ \begin{array}{l} f(t) = (t, 2t, 3t, \ldots), \\ g(t) = (t, t, t, \ldots), \\ h(t) = (t, \tfrac{1}{2}t, \tfrac{1}{3}t, \ldots). \end{array} \]
- In which topologies do the following sequences converge?
\[ \mathbf{w}_1 = (1, 1, 1, 1, \dots), \quad \mathbf{x}_1 = (1, 1, 1, 1, \dots), \]
\[ \mathbf{w}_2 = (0, 2, 2, 2, \dots), \quad \mathbf{x}_2 = (0, \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2}, \dots), \]
\[ \mathbf{w}_3 = (0, 0, 3, 3, \dots), \quad \mathbf{x}_3 = (0, 0, \tfrac{1}{3}, \tfrac{1}{3}, \dots), \]
\[ \mathbf{y}_1 = (1, 0, 0, 0, \dots), \quad \mathbf{z}_1 = (1, 1, 0, 0, \dots), \]
\[ \mathbf{y}_2 = (\tfrac{1}{2}, \tfrac{1}{2}, 0, 0, \dots), \quad \mathbf{z}_2 = (\tfrac{1}{2}, \tfrac{1}{2}, 0, 0, \dots), \]
\[ \mathbf{y}_3 = (\tfrac{1}{3}, \tfrac{1}{3}, \tfrac{1}{3}, 0, \dots), \quad \mathbf{z}_3 = (\tfrac{1}{3}, \tfrac{1}{3}, 0, 0, \dots). \]
Solution
The uniform topology is given by \(\bar\rho(\mathbf a,\mathbf b)=\sup_k\min\{\abs{a_k-b_k},1\}\), and \(\mathcal T_{\mathrm{prod}}\subseteq\mathcal T_{\bar\rho}\subseteq\mathcal T_{\mathrm{box}}\). A map into \(\mathbb R^\omega\) is product-continuous exactly when its coordinates are continuous, and a sequence converges in the product topology exactly when it converges coordinatewise. Continuity or convergence in a finer topology implies it in a coarser one. Let \(U=\prod_k(-\frac1{k^2},\frac1{k^2})\), a box neighborhood of \(\mathbf 0\). In (b), \(\mathbf w_n\) and \(\mathbf x_n\) have \(n-1\) leading zeros followed by the constant \(n\), respectively \(\frac1n\); \(\mathbf y_n\) has \(n\) entries \(\frac1n\) followed by zeros; \(\mathbf z_n\) has two entries \(\frac1n\) followed by zeros.
1. \(f\) is continuous \(\boxed{\text{in the product topology only}}\).
Its coordinates \(t\mapsto kt\) are continuous. For \(t\ne0\), \(\bar\rho(f(t),f(0))=\sup_k\min\{k\abs t,1\}=1\), so \(f\) is not uniformly continuous at \(0\), hence not box-continuous.
2. \(g\) and \(h\) are continuous \(\boxed{\text{in the product and uniform topologies, not in the box topology}}\).
Their coordinates \(t\mapsto t\) and \(t\mapsto t/k\) are continuous. \(\bar\rho(g(t),g(s))=\min\{\abs{t-s},1\}\) and \(\bar\rho(h(t),h(s))=\min\{\sup_k\abs{t-s}/k,1\}=\min\{\abs{t-s},1\}\). In the box topology, \(g^{-1}\bigl(\prod_k(-\frac1k,\frac1k)\bigr)=\{0\}\) and \(h^{-1}(U)=\{t:\abs t<\frac1k\text{ for all }k\}=\{0\}\), which are not open.
3. \((\mathbf w_n)\) converges to \(\mathbf 0\) \(\boxed{\text{in the product topology only}}\).
For each \(k\), the \(k\)-th coordinate of \(\mathbf w_n\) is \(0\) once \(n>k\). But \(\bar\rho(\mathbf w_n,\mathbf 0)=1\) for every \(n\), and the product limit \(\mathbf 0\) is the only possible limit in the finer Hausdorff topologies.
4. \((\mathbf x_n)\) and \((\mathbf y_n)\) converge to \(\mathbf 0\) \(\boxed{\text{in the product and uniform topologies, not in the box topology}}\).
\(\bar\rho(\mathbf x_n,\mathbf 0)=\bar\rho(\mathbf y_n,\mathbf 0)=\frac1n\to0\). The \(n\)-th coordinate of both \(\mathbf x_n\) and \(\mathbf y_n\) is \(\frac1n\ge\frac1{n^2}\), so neither sequence ever enters \(U\).
5. \((\mathbf z_n)\) converges to \(\mathbf 0\) \(\boxed{\text{in all three topologies}}\).
Let \(\prod_k(-a_k,a_k)\) be a box neighborhood of \(\mathbf 0\). The coordinates of \(\mathbf z_n\) beyond the second are \(0\), so \(\mathbf z_n\) lies in it once \(\frac1n<\min\{a_1,a_2\}\). Every box neighborhood of \(\mathbf 0\) contains one of this form.
6. Q.E.D.
Steps 1 and 2 answer (a), and steps 3, 4, and 5 answer (b).