Consider the space $C[0,\infty)$ of continuous functions on $[0,\infty)$. Consider the following two topologies.

The first topology, denoted $\tau_1$ is generated by cylinder sets, i.e. sets of the form: $\{\omega\in C[0,\infty) \text{ such that }(\omega_{t_1},..\omega_{t_2})\in A_1\times...\times A_n\}$, where $A_1,..A_n$ are open sets in $\mathbb{R}$.

The second topology, denoted $\tau_2$ is the metric topology induced by the following metric: $d(\omega_1,\omega_2) = \sum_{n=1}^\infty\frac{1}{2^n}\sup_{0\leq t \leq n}(|\omega_1(t)-\omega_2(t)|\wedge 1)$.

What I'm trying to answer is whether the two topologies coincide. I have managed to show that $\tau_1\subseteq\tau_2$. I'm not sure how to proceed with the other direction.

Any hint or direction or reference material would be greatly appreciated.


  • 2
    $\begingroup$ They don't coincide $\endgroup$ – Bananach Apr 7 at 19:53
  • $\begingroup$ @Bananach Thank you! $\endgroup$ – Viet Dang Apr 7 at 20:01

They do not coincide. For instance, pointwise convergence implies convergence in $\tau_1$, but not in $\tau_2$.

For example, let $h(x)$ be $1$ on $[2; 3]$ and $0$ on $[0; 1] \cup [4; \infty)$. Let $f_n(x) = h(\frac{x}{n})$. Then $f_n \to_{\tau_1} 0$ (as for any $x$ for all sufficiently large $n$ we have $f_n(x) = 0$), but $d(f_n, 0) \geqslant \frac{1}{2}$ for all $n$.

  • $\begingroup$ That was very helpful, thank you! $\endgroup$ – Viet Dang Apr 7 at 20:03
  • $\begingroup$ @VietDang So, why not accepting this answer? $\endgroup$ – Jochen Apr 9 at 7:49
  • $\begingroup$ @Jochen I did now. I'm not very familiar with the site. $\endgroup$ – Viet Dang Apr 16 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.