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(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:

  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) 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?

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  • $\begingroup$ I was curious enough to go off wandering, and, well, I guess now I know that 'viz.' for 'videlicet' is not just some bizarre misspelling of 'z' for 'd' but a corruption of vi…ꝫ, with the ꝫ short for 'et'. $\endgroup$
    – LSpice
    Apr 4, 2022 at 14:42
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    $\begingroup$ Yup! I seem to be very lonely, so far, in writing “viꝫ” for “videlicet”, but I hope the trend of actually using this nice Unicode character will catch! $\endgroup$
    – Gro-Tsen
    Apr 4, 2022 at 15:01

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I think the following is a counter-example for the equivalence of (2) and (3):

Consider the Baire space $\mathbf N^\mathbf N$ 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.

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