Locally convex topology vs compact open topology Consider the set of seminorms $p_n$ on $X = C(\mathbb R)$,
$$p_n(f) = \sup_{[-n, n]} |f(x)|.$$
Consider the topology on $C(\mathbb R)$, for which sets of the form
$$V_{f_0, n, \varepsilon} = \{f \in C(\mathbb R) \mid p_n(f-f_0) < \varepsilon\},$$
where $f_0 \in C(\mathbb R)$, $n \in \mathbb N$, $\varepsilon > 0$, are sub-base.
Show that this topology equals to the compact-open topology on $C(\mathbb R)$, that is, the topology with sub-base consisting of sets
$$W_{K, U} = \{f \in C(\mathbb R) \mid f(K) \subset U\},$$
where $K \subset \mathbb R$ is compact and $U \subset \mathbb R$ is open.
I honestly cannot even think of an approach for neither of the two directions. My lecturer just showed this as an example for some other concept and did not prove nor even explained the intuition.
 A: This is a special case of some very general results about uniform spaces and compact sets. See John Kelley's General Topology for that level of generality. (Maybe there is a quicker way than the sketch I provide below, but I'm not aware of it at the moment.)
You can start by considering the fact that $f(K) \subseteq U$ implies that there is some $\varepsilon > 0$ so that $\bigcup_{x \in K} B(f(x),\varepsilon) \subseteq U$. Indeed, for each $x \in K$, let $\varepsilon_x > 0$ be so that $B(f(x),3\varepsilon_x) \subseteq U$. By compactness of $f(K)$, there are is a finite family $x_1, \ldots, x_k$ so that $f(K) \subseteq \bigcup_{j=1}^k B(f(x_k),\varepsilon_{x_k})$. Now, let $\varepsilon = \min\{\varepsilon_{x_\ell} : 1 \leq \ell \leq k\}$. For $x \in K$, we want to show that $B(f(x),\varepsilon) \subseteq U$. So let $y \in B(f(x),\varepsilon)$. Also, let $x_\ell$ be so that $f(x) \in B(f(x_\ell),\varepsilon_{x_\ell})$. Now $$|y-f(x_\ell)| \leq |y-f(x)| + |f(x)-f(x_\ell)| < \varepsilon + \varepsilon_{x_\ell} \leq 2 \varepsilon_{x_\ell} < 3\varepsilon_{x_\ell}.$$ That is, $y \in B(f(x_\ell),3\varepsilon_{x_\ell}) \subseteq U$. So this $\varepsilon > 0$ can serve as a way to make sure other functions are "close enough" point-wise to $f$ on $K$. You can combine this with the fact that every compact $K$ is contained in some $[-n,n]$. That should establish that every $W_{K,U}$ can be written as the union of $V_{f,n,\varepsilon}$.
For the other direction, we sort of "cut up" the $[-n,n]$ in such a way that we make things fit in correctly. The reason this move is necessary, is because $x$ and $-x$ are both in $W_{[-2,2],(-2.1,2.1)}$ but aren't uniformly close to each other on $[-2,2]$.
Another thing to take note of is that, at least for a compact set $K$, $$\sup\{|f(x) - g(x)| : x \in K\} < \varepsilon \iff (\forall x \in K) |f(x) - g(x)| < \varepsilon.$$
So, suppose $f \in V_{f_0,n,\varepsilon}$. Check that (by considering $(\varepsilon - p_n(f-f_0))$ and the triangle inequality for $p_n$) we can find $\delta>0$ so that $V_{f,n,\delta} \subseteq V_{f_0,n,\varepsilon}$. By compactness of $f([-n,n])$, we can find $x_1 , \ldots , x_k \in [-n,n]$ so that $f([-n,n]) \subseteq \bigcup_{j=1}^k B(f(x_j),\delta/3)$. Then let $$K_j = [-n,n] \cap f^{-1}\left(\overline{B}(f(x_j),\delta/3)\right),$$ where $\overline{B}$ denotes the closed disk. Notice that each $K_j$ is compact and let $U_j = B(f(x_j) , \delta/2).$ The final claim is that $$f \in \bigcap_{j=1}^k W_{K_j,U_j} \subseteq V_{f_0,n,\varepsilon}.$$ The elemental relation is evident so let's move on to the subset containment. Let $g \in \bigcap_{j=1}^k W_{K_j,U_j}$ and $x \in [-n,n]$. Then $x \in K_j$ for some $j$ which means that $g(x) \in B(f(x_j),\delta/2)$. Moreover, $f(x) \in \overline{B}(f(x_j),\delta/3)$. Then $$|f(x) - g(x)| \leq |f(x) - f(x_j)| + |f(x_j) - g(x)| \leq \delta/3 + \delta/2 < \delta.$$ Hence, $g \in V_{f,n,\delta} \subseteq V_{f_0,n,\varepsilon}$.
