Your intuition is right for the first question. Let $y_n = x_n + (1 - 2^{-n})\epsilon$; then clearly $y = (y_n) \in U(x,\epsilon)$. But no ball around $y$ in the uniform metric is contained in $U(x,\epsilon)$: if the radius of such a ball is $\delta$ and $\delta > 2^{-n}\epsilon$ for some $n$, then $B_\delta(y)$ contains, for example, $z = (z_k)$ where
$$z_k = \begin{cases} y_k & k \ne n \\ y_n + 2^{-n}\epsilon & k = n, \end{cases}$$
which is not in $U(x,\epsilon)$.
The second question is asking you to check two things. The first is that every $U(x,\delta)$ for $\delta < \epsilon$ is contained in $B_\epsilon(x)$. That is, every point $y$ in $U(x,\delta)$ has $\rho(x,y) < \epsilon$. But in fact, $\rho(x,y) \le \delta$, since $|y_n - x_n| < \delta$ for each $n$. (Make sure you understand why this is $\le \delta$ and not $< \delta$ -- this is the same principle behind my answer to the first question.) The second thing to check is that every $y$ with $\rho(x,y) < \epsilon$ is contained in some $U(x,\delta)$. If $\rho(x,y) = \epsilon' < \delta < \epsilon$, then you can easily check that $y$ is contained in $U(x,\delta)$.
Here's some 'philosophy' that may be helpful. Membership in these box-topology-style open sets like $U(x,\epsilon)$ is determined by each coordinate individually, but membership in uniform open sets is determined by the behavior of the entire sequence $(y_n)$, since you're taking a supremum.
To see it a bit more generally: we can define a topology on a space $X$ by specifying for each $x \in X$ a non-empty collection $\mathcal{B}_x$ of subsets of $X$ that obey the following axioms:
- $\forall x \in X: \forall B \in \mathcal{B}_x: x \in B$.
- $\forall x \in X: \forall B_1, B_2 \in \mathcal{B}_x: \exists B_3 \in \mathcal{B}_x: B_3 \subseteq B_1 \cap B_2$.
- $\forall x \in X: \forall B \in \mathcal{B}_x: \exists O \subset X: O \text { is open and } x \in O \subset B$.
Here a subset $O \subseteq X$ is called "open" when $\forall x \in O: \exists B \in \mathcal{B}_x: B \subseteq O$.
If the collections $\mathcal{B}_x$ satisfy these axioms, the collection of "open" subsets (as defined above) does indeed form a topology $\mathcal{T}$ (as the name suggests) and the collections $\mathcal{B}_x$ form a local base at $x$ for the topology $\mathcal{T}$.
Now in the context of the set $Y^X$, where $X$ is any space (set, even) and $(Y,d)$ any metric space, we can define for each $f$, the collection $\mathcal{B}_f = \{B(f, \epsilon): \epsilon > 0 \}$, and verify that these obey the axioms. Axiom (1) is clear, as $d(f(x), f(x)) = 0 < \epsilon$, etc.
Axiom (2) is also clear, as $B(f, \epsilon_1) \cap B(f, \epsilon_2) = B(f, \min(\epsilon_1, \epsilon_2))$.
Axiom (3) is more interesting: we claim that $B(f, \epsilon)$ is "open" for every $\epsilon > 0$ and any $f$: suppose $g \in B(f, \epsilon)$, so $s := \sup_{x \in X} d(f(x), g(x)) < \epsilon$, so $t = \frac{\epsilon - s}{2} > 0$. Then $B(g, t) \subseteq B(f, \epsilon)$: take $h \in B(g,t)$, then for any $x \in X$: $d(h(x), f(x)) \le d(h(x), g(x)) + d(g(x), f(x)) \le t + s$, so $\sup_{x \in X} d(f(x), h(x)) \le t + s < (\epsilon -s ) + s = \epsilon$, and this implies that $h \in B(f, \epsilon)$ and the inclusion has been shown. As $g \in B(f, \epsilon)$ was arbitary, we have shown that $B(f, \epsilon)$ is itself open, so (3) is now trivial.
Note that for this we need no topology on $X$ at all. We do know now that the $B(f, \epsilon)$ form a local base for the topology at $f$ (which is a bit more informative that that they are in a subbase). This is the topology of uniform convergence (which you call the uniform topology).
Quite similarly, when $X$ is a topological space, we can define another set of collections $\mathcal{B}'_f = \{B_K(f, \epsilon): \epsilon> 0, K \subset X \text{ compact} \}$ and see that these also satisy the axioms (1)-(3).
The fact that (1) holds is almost the same as above, for (2) we note that $$B_{K_1 \cup K_2}(f, \min(\epsilon_1,\epsilon_2)) \subseteq B_{K_1}(f, \epsilon_1) \cap B_{K_2}(f, \epsilon_2)\text{,}$$
where we use that the collection of compact subsets of $X$ is closed under finite unions.
As to (3), the proof that each $B_K(f, \epsilon)$ is open is almost literally the same as above, except that we take the $\sup$ over members of $K$ only.
So again, the collections $\mathcal{B}'_f$ form a local base at $f$ for the so-called topology of uniform convergence on compacta (aka as the compact convergence topology).
Now we only need to remark that almost trivally (as $\sup_{x \in K} d(f(x), g(x)) \le \sup_{x \in X} d(f(x), g(x))$), $B(f, \epsilon) \subseteq B_K(f, \epsilon)$ for all $K \subset X$, so when $O$ is open in the topology of uniform convergence on compacta, and $f \in O$, it contains some $B_K(f, \epsilon) \subset O$, and so it also contains $B(f, \epsilon)$, and $f$ is an interior point for the topology of uniform convergence. So $O$ is open in that topology as well. So the uniform topology is a superset of the topology of compact convergence.
We could also take the collection of finite subsets and get the pointwise topology instead of the compact convergence topology, and as all finite substes are compact, we trivally have that the compact convergence topology is a superset of the pointwise topology.
So $\mathcal{T}_{pw} \subseteq \mathcal{T}_{cc} \subseteq \mathcal{T}_u$, where the last two coincide when $X$ is compact itself, and the first two when e.g. the only compact subsets of $X$ are the finite ones.
Best Answer
You appear to be trying to prove the wrong thing. To show that the box topology is at least as fine as the uniform topology, you need to show that $\mathscr{T}_u\subseteq\mathscr{T}_b$, and if you want to show that it’s strictly finer, you have to show further that $\mathscr{T}_b\setminus\mathscr{T}_u\ne\varnothing$.
Let $U\in\mathscr{T}_u$; if $U=\varnothing$, then certainly $U\in\mathscr{T}_b$, so assume that $U\ne\varnothing$, and let $x\in U$; there is an $\epsilon(x)\in(0,1)$ such that $B_{2\epsilon(x)}^u(x)\subseteq U$. Let
$$V_x=\big(x-\epsilon(x),x+\epsilon(x)\big)^{\Bbb N}\;;$$
clearly $V_x\in\mathscr{T}_b$, and $d_u(x,y)\le\epsilon<2\epsilon$ for each $y\in V_x$, so $x\in V_x\subseteq U$. Thus,
$$U=\bigcup_{x\in U}V_x\in\mathscr{T}_b\;,$$
and $\mathscr{T}_u\subseteq\mathscr{T}_b$.
To show that $\mathscr{T}_b\setminus\mathscr{T}_u\ne\varnothing$, let
$$\begin{align*} U&=\prod_{n\in\Bbb N}\left(-\frac1{n+1},\frac1{n+1}\right)\\ &=(-1,1)\times\left(-\frac12,\frac12\right)\times\left(-\frac13,\frac13\right)\times\ldots\;; \end{align*}$$
I’ll leave it to you to verify that $U\notin\mathscr{T}_u$.