(I asked this on MSE a week ago, but did not get any answers there, so I'm trying here.)

Let $X$ be a topological space. I will define four ways in which a sequence $(f_n)$ of *continuous* functions $X \to \mathbb{R}$ might converge to a *continuous* function $f\colon X \to \mathbb{R}$, four notions intermediate between pointwise convergence (labeled (1) below) and uniform convergence (labeled (4) below). (Note: I am listing all four below in order to justify why I care about (2) and (3), but only (2) and (3) are involved in the actual question.) Specifically:

$f_n$ tends to $f$ pointwise, namely: for all $x\in X$ we have $f_n(x) \to f(x)$ (with no assumption of uniformity of any kind), viꝫ., for all $\varepsilon>0$ and all $x\in X$ there is $n_0$ such that when $n\geq n_0$ we have $|f_n(x)-f(x)|<\varepsilon$.

We now demand that (for $\varepsilon>0$ fixed), $n_0$ take the same value on each element of an open cover $(U_i)_{i\in I}$ of $X$. More precisely: for all $\varepsilon>0$ there is a covering $(U_i)_{i\in I}$ of $X$ by open sets such that for all $i\in I$ there is $n_0$ such that when $x \in U_i$ and $n\geq n_0$ we have $|f_n(x)-f(x)|<\varepsilon$.

We now demand that the same open cover $(U_i)_{i\in I}$ work for every $\varepsilon>0$; this is the same as demanding that $f_n\to f$ uniformly on each $U_i$, viꝫ. there is a covering $(U_i)_{i\in I}$ of $X$ by open sets such that for all $\varepsilon>0$ and all $i\in I$ there is $n_0$ such that when $x \in U_i$ and $n\geq n_0$ we have $|f_n(x)-f(x)|<\varepsilon$. (I suppose we could say that $f_n \to f$ “locally uniformly”, but I'm not sure whether this is standard terminology.)

We now demand that the covering consist of just $X$, i.e., that $f_n \to f$ uniformly, viꝫ. there for all $\varepsilon>0$ there is $n_0$ such that when $x \in X$ and $n\geq n_0$ we have $|f_n(x)-f(x)|<\varepsilon$.

To summarize:

$\forall \varepsilon>0. \forall x\in X. \exists n_0\in \mathbb{N}. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$

$\forall \varepsilon>0. \exists (U_i) \text{ open covering}. \forall i\in I. \exists n_0\in \mathbb{N}. \forall x\in U_i. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$

$\exists (U_i) \text{ open covering}. \forall \varepsilon>0. \forall i\in I. \exists n_0\in \mathbb{N}. \forall x\in U_i. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$

$\forall \varepsilon>0. \exists n_0\in \mathbb{N}. \forall x\in X. \forall n≥n_0. |f_n(x)-f(x)|<\varepsilon$

It's clear that $(4)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1)$.

A counterexample showing that (3) does not imply (4) is given by $X = \mathbb{R}$ and $f_n(x) = \exp(-(x-n)^2)$, which converges in sense (3) but not uniformly (4) toward $f = 0$.

A counterexample showing that (1) does not imply (2) is given by $X = [0,1]$ and $f_n(x) = \max(0, \min(nx, 2-nx))$ (graphs here) which converge pointwise (1) toward $f = 0$ but there is no neighborhood $V$ of $0$ such that $\exists n_0\in \mathbb{N}. \forall x\in V. \forall n\geq n_0. |f_n(x)|<\frac{1}{2}$ so (2) does not hold.

**Question:** Does (2) have a standard name? Are (2) and (3) perhaps in fact equivalent? If yes, what is a proof? If no, what is a counterexample?