(I asked this [on MSE](https://math.stackexchange.com/questions/4268290/relation-between-two-notions-intermediate-between-pointwise-convergence-and-u) 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:


1. $f\_n$ 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 $|f\_n(x)-f(x)|<\varepsilon$.
2. 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 $|f\_n(x)-f(x)|<\varepsilon$.
3. 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.)
4. 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:


1. $\forall \varepsilon>0. \forall x\in X. \exists n\_0\in \mathbb{N}. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
2. $\forall \varepsilon>0. \exists (U\_i) \text{ open covering}. \forall i\in I. \exists n\_0\in \mathbb{N}. \forall x\in U\_i. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
3. $\exists (U\_i) \text{ open covering}. \forall \varepsilon>0. \forall i\in I. \exists n\_0\in \mathbb{N}. \forall x\in U\_i. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
4. $\forall \varepsilon>0. \exists n\_0\in \mathbb{N}. \forall x\in X. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$


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](https://twitter.com/gro_tsen/status/1444595905374658566)) 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?



Relation between two notions intermediate between “pointwise convergence” and “uniform convergence”
I think the following is a counter-example for the equivalence of (2) and (3):


Consider the [Baire space](https://en.wikipedia.org/wiki/Baire_space_(set_theory)) $\mathbf N^\mathbf N$ and let $f\_n$ be the following function: If your sequence $x$ starts with $k$ zeroes followed by the entry $n$ then $f\_n(x) = 1/(k+1)$, otherwise $f\_n(x) = 0$.


The $f\_n$ are continuous: away from the sequence that is constantly 0, they are even locally constant. At $0^\infty$ you can easily check continuity using cylinder sets.


The $f\_n$ satisfy (2): If $\epsilon > 1/(k+1)$, we can take as our covering the $k$-cylinder sets: for $v \in \mathbf N^k$, let $U\_v$ be the set of sequences that start with $v$. If $v = 0^k$, then $\sup f\_n(U\_v) = 1/(k+1) < \epsilon$. For $v \not=0^k$, we can wait until n is larger than the first nonzero entry of $v$, and get $f\_n(U\_v) = 0$ for all $n \geq n\_0$ for some $n\_0$. So the $U\_v$ form the required covering.


The $f\_n$ do not satisfy (3): Some set of the open covering must contain $0^\infty$, and so it must contain a cylinder set $U\_v$ with $v = 0^k$ for some $k$. But $\sup f\_n(U\_v) = 1/(k+1)$ for all $n$, so you don't have local uniform convergence.


In my construction I heavily used that $\mathbf N^\mathbf N$ is not locally compact, I am curious whether (2) and (3) are equivalent for locally compact spaces.


