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