Consider the map \(h: \mathbb{R}^\omega \to \mathbb{R}^\omega\) defined in Exercise 8 of §19; give \(\mathbb{R}^\omega\) the uniform topology. Under what conditions on the numbers \(a_i\) and \(b_i\) is \(h\) continuous? a homeomorphism?
Solution
Here \(h(\mathbf x)=(a_ix_i+b_i)_{i\ge1}\) with \(a_i>0\), as in E-AWEWJ, and the uniform topology is given by \(\bar\rho(\mathbf x,\mathbf y)=\sup_i\min\{\abs{x_i-y_i},1\}\).
1. Translation \(\mathbf x\mapsto\mathbf x+\mathbf b\) is an isometry of \((\mathbb R^\omega,\bar\rho)\), so \(h\) is continuous, or a homeomorphism, if and only if \(\mathbf x\mapsto(a_ix_i)_i\) is.
\(\min\{\abs{(x_i+b_i)-(y_i+b_i)},1\}=\min\{\abs{x_i-y_i},1\}\) for every \(i\).
2. \(h\) is continuous if and only if \(\boxed{\sup_i a_i<\infty}\).
Suppose \(M=\sup_ia_i<\infty\), and let \(0<\varepsilon\le1\). If \(\bar\rho(\mathbf x,\mathbf y)<\min\{\varepsilon/M,1\}\), then \(\abs{x_i-y_i}<\varepsilon/M\) for every \(i\), so \(\abs{a_ix_i-a_iy_i}<\varepsilon\) and \(\bar\rho(h\mathbf x,h\mathbf y)\le\varepsilon\).
Suppose \((a_i)\) is unbounded, and let \(\delta\in(0,1)\). Choose \(k\) with \(a_k>1/\delta\), and let \(\mathbf y\) have \(k\)-th coordinate \(\delta/2\) and all other coordinates \(0\). Then \(\bar\rho(\mathbf y,\mathbf 0)=\delta/2<\delta\), but the \(k\)-th coordinates of \(h\mathbf y\) and \(h\mathbf 0\) differ by \(a_k\delta/2>\frac12\), so \(\bar\rho(h\mathbf y,h\mathbf 0)>\frac12\). Hence \(h\) is not continuous at \(\mathbf 0\).
3. \(h\) is a homeomorphism if and only if \(\boxed{\sup_ia_i<\infty\text{ and }\inf_ia_i>0}\).
Since every \(a_i>0\), \(h\) is bijective with inverse \(h^{-1}(\mathbf y)=\bigl((y_i-b_i)/a_i\bigr)_i\), a map of the same form with coefficients \(1/a_i\). By step 2, \(h\) is continuous if and only if \(\sup_ia_i<\infty\), and \(h^{-1}\) is continuous if and only if \(\sup_i(1/a_i)<\infty\), that is, \(\inf_ia_i>0\).
4. Q.E.D.
Steps 2 and 3 give the two conditions; by step 1 neither depends on \((b_i)\).