Here is Prob. 20 (a) in the book Topology by James R. Munkres, 2nd edition.

Consider the product, uniform, and box topologies on $\mathbb{R}^\omega$. In which of these topologies are the following functions from $\mathbb{R}$ to $\mathbb{R}^\omega$ continuous? $$f(t) = \left( t, 2t, 3t, \ldots \right),$$ $$g(t) = \left( t, t, t, \ldots \right),$$ $$h(t) = \left( t, \frac{t}{2}, \frac{t}{3}, \ldots \right)$$ for each $t \in \mathbb{R}$.

Let $U \colon= \prod_{n \in \mathbb{N}} U_n$ be a product topology basis element for $\mathbb{R}^\omega$; let $n_1, \ldots, n_k$ be the natural numbers such that $U_{n_i}$ is a proper open subset of $\mathbb{R}$ for each $i = 1, \ldots, k$; suppose $U_n = \mathbb{R}$ for all other $n \in \mathbb{N}$.

Suppose $t \in f^{-1}(U)$ so that $f(t) \in U$ and hence $$\pi_{n_i}\left(f(t) \right) = n_i t \in U_{n_i}$$ for each $i = 1, \ldots, k$. Here, for each $n \in \mathbb{N}$, the map $\pi_n \colon \mathbb{R}^\omega \to \mathbb{R}$ is defined by $$\pi_n(x) \colon= x_n \ \mbox{ for all } \ x \colon= \left( x_1, x_2, x_3, \ldots \right) \in \mathbb{R}^\omega.$$ Since $U_{n_i}$ is open in $\mathbb{R}$ under the usual topology, there exists a positive real number $\delta_i$ such that $$\left( \pi_{n_i}\left( f(t) \right) - \delta_i, \pi_{n_i}\left( f(t) \right) + \delta_i \right) = \left( n_i t - \delta_i , n_i t + \delta_i \right) \subset U_{n_i}$$ for each $i = 1, \ldots, k$. Let $$\delta \colon= \min \left\{ \frac{\delta_1}{n_1}, \ldots, \frac{\delta_k}{n_k} \right\}. $$ This $\delta$ is a positive real number, and if $s \in (t- \delta, t+\delta)$, then $$ t - \frac{ \delta_{n_i} }{ n_i } < s < t + \frac{ \delta_{n_i} }{ n_i },$$ which implies that $$n_i t - \delta_i < n_i s < n_i t + \delta_i,$$ which in turn implies that $$n_i s \in U_{n_i},$$ for each $i = 1, \ldots, k$, and hence $f(s) \in U$.

Thus, for each $t \in f^{-1}(U)$, there is a positive real number $\delta$ such that $$t \in (t- \delta, t + \delta) \subset f^{-1}(U),$$ showing that the inverse image $f^{-1}(U)$ is open for each basis element $U$ of the product topology on $\mathbb{R}^\omega$.

Now suppose that $t \in g^{-1}(U)$. Then, for each $i = 1, \ldots, k$, $$\pi_{n_i}\left( g(t) \right) = t \in U_{n_i}$$ and hence $$\left( t - \delta_i, t+\delta_i \right) \subset U_{n_i}$$ for some positive real number $\delta_i$. Let $$\delta \colon= \min \left\{ \delta_1, \ldots, \delta_k \right\}.$$ This $\delta$ is a positive real number, and if $s \in ( t-\delta, t+\delta )$, then $s \in \left(t-\delta_i, t + \delta_i \right)$ and hence $g(s) \in U$, showing that $g^{-1}(U)$ is open.

Finaly, if $t \in h^{-1}(U)$, then, for each $i = 1, \ldots, k$, $$ \pi_{n_i}\left( h(t) \right) = \frac{t}{n_i} \in U_{n_i}$$ and hence $$\left( \frac{t}{n_i} - \delta_i , \frac{t}{n_i} + \delta_i \right) \subset U_{n_i}$$ for some positive real number $\delta_i$. Let's take $$\delta \colon= \min \left\{ n_1 \delta_1, \ldots, n_k \delta_k \right\}.$$ This $\delta $ is a positive real number, and if $s \in (t - \delta , t + \delta )$, then $$t- n_i \delta_i < s < t + n_i \delta_i$$ and hence $$ \frac{t}{n_i} - \delta_i < \frac{s}{n_i} < \frac{t}{n_i} + \delta_i $$ or $$ \pi_{n_i}\left( h(s) \right) \in \left( \frac{t}{n_i} - \delta_i, \frac{t}{n_i} + \delta_i \right) \subset U_{n_i}$$ for each $i = 1, \ldots, k$, which implies that $s \in h^{-1}(U)$, from which it follows that $h^{-1}(U)$ is open in $\mathbb{R}$.

Have I been able to get these proofs right?

Now for the uniform topology.

Let's take a real number $\varepsilon$ such that $0 < \varepsilon < \frac 1 2$. Let $t \in \mathbb{R}$. If $s \in \mathbb{R}$ such that $s \neq t$, then $$\tilde{\rho}\left( f(s), f(t) \right) = \sup \left\{ \ \min \left\{ n \vert s-t\vert, 1 \right\} \ \colon \ n \in \mathbb{N} \ \right\} = 1 > \varepsilon,$$ from which it follows that $f$ is not continuous (at any point of $\mathbb{R}$).

Now let us take a real number $\delta$ such that $0 < \delta \leq \varepsilon$. Then, for any points $s, t \in \mathbb{R}$ such that $\vert s-t \vert < \delta$, we have $$\tilde{\rho}\left( g(s), g(t) \right) = \sup \left\{ \ \min \left\{ \vert s-t \vert, 1 \right\} \ \colon \ n \in \mathbb{N} \right\} = \vert s- t \vert < \varepsilon,$$ and hence it follows that $g$ is (uniformly) continuous.

Now let's choose $\varepsilon$ and $\delta$ as above. Then, for any $s, t \in \mathbb{R}$ such that $\vert s-t\vert < \delta$, we have $$\tilde{\rho}\left( h(s), h(t) \right) = \sup \left\{ \ \min \left\{ \frac{\vert s-t \vert}{n}, 1 \right\} \ \colon \ n \in \mathbb{N} \ \right\} = \min \left\{ \vert s-t\vert, 1 \right\} = \vert s-t\vert < \varepsilon,$$ from which it follows that $h$ is (uniformly) continuous.

Have I been able to come up with the right answer in each case? If so, have I been able to get these proofs right as well?

For the product topology, we can also use Theorem 19.6 in Munkres. Here I have attempted to show the continuity of $f$, $g$, and $h$ directly.

And, the box topology on $\mathbb{R}^\omega$ is the one havaing as a basis all sets of the form $$ (a_1, b_1) \times (a_2, b_2) \times (a_3, b_3) \times \cdots, $$ where $\left(a_i \right)_{i \in \mathbb{N}} \in \mathbb{R}^\omega$ and $\left(b_i \right)_{i \in \mathbb{N}} \in \mathbb{R}^\omega$ are such that $a_i < b_i$ for each $i = 1, 2, 3, \ldots$, and $(a_i, b_i)$ denotes the segment (i.e. open interval) with $a_i$ as the left endpoint and $b_i$ as the right endpoint.

The functions $f$, $g$, and $h$ are not continuous when $\mathbb{R}^\omega$ is given the box topology. The inverse image under each of $f$, $g$, and $h$ of the basis element $$B \colon= \left( -1, 1 \right) \times \left(-\frac{1}{2^2}, \frac{1}{2^2} \right) \times \left( -\frac{1}{3^2}, \frac{1}{3^2} \right) \times \cdots, $$ for example, contains the point $t = 0$ since $$f(0) = g(0) = h(0) = (0, 0, 0, \ldots) \in B.$$ So, in order for this inverse image to be open in $\mathbb{R}$ with the usual topology, there must be an open interval $\left(-\delta_f, \delta_f \right)$, $\left( -\delta_g, \delta_g \right)$, and $\left( -\delta_h, \delta_h \right)$, for some positive real numbers $\delta_f$, $\delta_g$, and $\delta_h$, respectively, such that $$ \left(-\delta_f, \delta_f \right) \subset f^{-1}(B),$$ $$\left( -\delta_g, \delta_g \right) \subset g^{-1}(B), $$ $$\left( -\delta_h, \delta_h \right) \subset h^{-1}(B). $$ In particular, we must have $$f\left( \frac{\delta_f}{2} \right) \in B,$$ $$g\left( \frac{\delta_g}{2} \right) \in B,$$ $$h\left( \frac{\delta_h}{2} \right) \in B.$$ So, for each $n \in \mathbb{N}$, we have $$\frac{n \delta_f}{2} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), \ \frac{\delta_g}{2} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), \ \frac{\delta_h}{2n} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), $$ and hence $$n^3 < \frac{2}{ \delta_f } \ \mbox{ for all} \ n \in \mathbb{N},$$ $$n^2 < \frac{2}{ \delta_g } \ \mbox{ for all} \ n \in \mathbb{N},$$ $$n < \frac{2}{ \delta_h } \ \mbox{ for all} \ n \in \mathbb{N},$$ each of which is impossible.

Am I right?

That $g$ is not continuous was also shown by Munkres himself in Example 2, Sec. 19 on page 117.


1 Answer 1


Yes, your proofs are correct!

I think they could be more streamlined and readable, though. Due to the simplicity of the functions involved, it's easy to compute $v^{-1}(U)$ explicitly for $v = f, g, h$ and determine if the result is open in $\mathbb R$ rather than picking an arbitrary point $t \in v^{-1}(U)$ and then some $\delta> 0$ with $(t-\delta, t+\delta) \subseteq v^{-1}(U)$ to show that $v^{-1}(U)$ is open. In either case, it should first be justified continuity of $v$ is equivalent to "$v^{-1}(U)$ is open when $U$ is a basis element" (rather than an arbitrary open set) because we can always write $U$ as a union $U = \bigcup_\alpha U_\alpha$ of basis elements, and since inverse imeages preserve unions, $v^{-1}\left( \bigcup_\alpha U_\alpha \right) = \bigcup_\alpha v^{-1}(U_\alpha)$, the latter of which is open if $v^{-1}(U_\alpha)$ is open for basis elements $U_\alpha$.

By defining $v_n = \pi_n \circ v$, we can write $v \colon \mathbb R \to \mathbb R^\omega$ as $v = (v_1, v_2, \dots)$, where $v_n \colon \mathbb R \to \mathbb R$, in the sense that $$ v(t) = \big( v_1(t), v_2(t), \dots \big). $$ Substituting $v = f, g, h$, \begin{align*} f_n(t) = nt,\quad g_n(t) = t,\quad h_n(t) = \frac{t}{n}. \end{align*}

We can consider the product and box topologies simultaneously by taking a general (nonempty) basis element $$ U = (a_1, b_1) \times (a_2, b_2) \times \cdots, $$ where $-\infty \leq a_n < b_n \leq \infty$ for all $n \in \mathbb N$. The product topology just has the stipulation that $(a_n, b_n) \neq (-\infty, \infty) = \mathbb R$ only for a finite subset of indices $n_1, \dots, n_k \in \mathbb N$.

To compute $v^{-1}(U)$, note that $$ t \in v^{-1}(U) \iff v(t) \in U \iff \left[ v_n(t) \in (a_n, b_n) \quad\forall n \in \mathbb N \right]. $$ Hence

  • $t \in f^{-1}(U) \iff nt \in (a_n, b_n) \quad \forall n \iff t \in \left( \tfrac{a_n}{n}, \tfrac{b_n}{n} \right) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( \tfrac{a_n}{n}, \tfrac{b_n}{n} \right)$. The latter set is open in the product topology, being the intersection of only a finite number of open sets (since $(a_n, b_n) = \mathbb R$ for all but finitely many $n$), and hence $f$ is continuous in the product topology. However, that set is not always open in the box topology (take $a_n = -1$ and $b_n = 1$ for all $n$ to see $f^{-1}(U) = \{0\}$, which isn't open), and hence $f$ is not continuous in the box topology.
  • $t \in g^{-1}(U) \iff t \in (a_n, b_n) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( a_n, b_n \right)$. By the same reasoning as for $f$, $g$ is continuous in the product topology. In the box topology, taking $\left( a_n, b_n \right) = \left( -\tfrac{1}{n}, \tfrac{1}{n} \right)$ shows that $g$ is not continuous.
  • $t \in h^{-1}(U) \iff \tfrac{t}{n} \in (a_n, b_n) \quad \forall n \iff t \in \left( n a_n, n b_n \right) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( n a_n, n b_n \right)$. By the same reasoning, $h$ is continuous in the product topology. In the box topology, take $a_n = -\tfrac{1}{n^2}$ and $b_n = \tfrac{1}{n^2}$ to see that $h$ is not continuous.

For the uniform topology, I wouldn't change a thing!

  • 2
    $\begingroup$ @Leucippus If the proof is correct, what else is there to say? $\endgroup$ Jun 11, 2016 at 20:23
  • $\begingroup$ @StanCoreyCarter What is presented here is a comment and not an answer defined in the rules and regs of the site. Although correct, it doesn't satisfy what and "answer" is. $\endgroup$
    – Leucippus
    Jun 11, 2016 at 20:59
  • $\begingroup$ @StanPalasek I'm curious to know more about your undergrad degree, esp. the analysis, topology, and functional analysis courses at Princeton. Can we connect outside of Math SE as well, please? If you agree too, then my e-mail is saaqib1978@yahoo.co.in, and my mobile phone number (for WhatsApp, Skype, and Viber) is +92-334-541-7958. My Skype ID is saaqib.mahmood. Regards. $\endgroup$ Oct 19, 2016 at 17:40
  • $\begingroup$ @JonWarneke can you please take time having a look at my edit? $\endgroup$ Oct 19, 2016 at 17:56
  • $\begingroup$ Your edit looks good! $\endgroup$ Oct 19, 2016 at 18:00

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