Does locally uniform convergence define a topology I am TAing an undergraduate analysis course and got stumped on this question. Let $X$ be a topological space, and define a notion of sequential convergence on $C(X,\mathbb R)$ as follows. We say that $f_n\to f$ locally uniformly if for all $x\in X$, there exists a neighborhood $U\ni x$ such that $f_n\to f$ uniformly on $U$.
Is this induced by a topology? I would like to define a neighbourhood basis similarly to the case of compact convergence, by
$$
\mathscr{U}_{f,U,\epsilon} = \{ g\in C(X)\, :\, \sup_U|f-g| <\epsilon\}
$$
where $U\subset X$ is any open, but without restrictions on the size of $U$, this will simply give us uniform convergence again.
However, I'm also struggling to construct a counter example of the type used to show that almost everywhere convergence is not topologizable.
Of course for $X$ locally compact this question is trivial, so a potential counterexample would need a more careful choice of space $X$.
 A: I'm answering this in case anyone else arrives at this question. Write $(X,\tau_X)$ for the topology of $X$, and we define a topology $\tau_{LU}$ of locally uniform convergence on $C(X,\mathbb R)$ as follows.
Let $\mathcal B\subset \tau_X$ be a neighborhood basis of $(X,\tau_X)$ which is closed under finite unions. For any $f\in C(X,\mathbb R)$, any $U\in\mathcal B$, and any $\varepsilon>0$, we define the open
$$
\mathscr U_{f,U,\varepsilon} := \{g\in C(X):\sup_U|f-g|<\varepsilon\}.
$$
We now show that the family $\{\mathscr U_{f,U,\varepsilon}\}_{f\in C(X),U\in\mathcal B,\varepsilon>0}$ forms a neighborhood basis for $C(X)$. Suppose $g\in\mathscr U_{f_1,U_1,\varepsilon_1}\cap \mathscr U_{f_2,U_2,\varepsilon_2}$. Setting $\varepsilon_i' = \varepsilon_i - \sup_{U_i}|f_i-g|>0$ and letting $\varepsilon:=\min\{\varepsilon_1',\varepsilon_2'\}$, we get that
$$
g \in \mathscr U_{g,U_1\cup U_2,\varepsilon}\subset \mathscr U_{f_1,U_1,\varepsilon_1}\cap \mathscr U_{f_2,U_2,\varepsilon_2}.
$$
We call the resulting topology $\tau_{\mathcal B}$.
We can now define the topology
$$
\tau_{LU} = \bigcap_{\mathcal B}\tau_{\mathcal B}
$$
where we index over neighborhood bases $\mathcal B\subset\tau_X$ that are closed under finite unions. This gives the desired topology.
