Let $f$, $g$, $h \colon \mathbf{R} \to \mathbf{R}^\omega$ be defined by

$$\begin{align*} f(t)&:=(t,2t,3t,\ldots),\\\\ g(t)&:=(t,t,t,\ldots),\\\\ h(t)&:=\left(t,\frac{t}{2},\frac{t}{3},\ldots\right) \end{align*}$$

for all $t \in \mathbf{R}$. How to determine whether these functions are continuous relative to the uniform topology on $\mathbf{R}^\omega$? We of course assume that $\mathbf{R}$ is given the usual topology.



  1. Let $U=\{x\in\Bbb R^\omega:\|x\|<1\}$, where $\|x\|=\sup_{n\in\omega}|x_n|$. $U$ is open in $\Bbb R^\omega$. (Why?) What is $f^{-1}[U]$? Is it open in $\Bbb R$?

  2. Show that for any $s,t\in\Bbb R$, $\|g(s)-g(t)\|=|s-t|$; use this to show that $g$ is continuous. In fact you can use it to show that $g$ is even a homeomorphism.

  3. Show that for any $s,t\in\Bbb R$, $\|h(s)-h(t)\|=|s-t|$.

  • $\begingroup$ please note that here we're using the uniform metric on $\mathbb{R}^\omega$ which is defined as follows: $$\tilde{\rho}(x,y) \colon= \sup\{ \min(\vert x_n - y_n \vert, 1) : n = 1, 2, 3, \ldots\} \forall x \colon= (x_n)_{n \in \mathbb{N}}, y= (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega$$. So do you think you should revise your answer? $\endgroup$ – Saaqib Mahmood Apr 9 '15 at 15:31
  • $\begingroup$ @Saaqib: It’s the same thing, just a different notation: $\|x-y\|$, as I defined it, is exactly the same as your $\tilde{\rho}(x,y)$. $\endgroup$ – Brian M. Scott Apr 9 '15 at 16:45

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