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I am really struggling with the differences between convergence in the product topology and convergence in the box topology. More specifically, I have some doubts concerning the definitions of those concepts.

What I have got so far is the following definition (where $\mathcal{N}_f$ denotes a nhood of $f$), where it seems that superficially they do not look that different:

$f_n \to f \in \mathbb{R}^X$ converges in the ***-topology $\Longleftrightarrow \forall x \in X \ \forall V \in \mathcal{N}_f \ \exists N(x) \geq 1 : \forall n \geq N, \ f_n \in V$

where the "***" means that the definition works for both the product AND the box topology, and the only difference lies in the form that the $V \in \mathcal{N}_f$ have (i.e. the meaning of being an open set in the product and in the box).

Is this intuition correct?

Any feedback is most welcome.
Thank you for your time.

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2 Answers 2

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I will try to give you some help with the general definitions.

  • You wrote: "where $\mathcal{N}_f$ denotes a nhood of $f$". Each $V\in\mathcal{N}_f$ is a neighborhood of $f$. ($\mathcal{N}_x$ is some neighborhood base at $f$.)

  • Your general definition of $f_n\to f$ $$\forall x \in X \ \forall V \in \mathcal{N}_f \ \exists N(x) \geq 1 : \forall n \geq N, \ f_n \in V$$ is not correct. The general definition for convergence of a sequence $x_n\to x$ in a topological space is $$(\forall U\in\mathcal{N}_x)(\exists N)(\forall n\geq N): x_n\in U.$$ So, the general definition for convergence of a sequence $f_n\to f$ in $\mathbb{R}^X$ is $$(\forall U\in\mathcal{N}_f)(\exists N)(\forall n\geq N): f_n\in U.$$

You are right that the difference is the meaning of an open set in each topology. If you specify $\mathcal{N}_f$ for each topology (product/box/uniform) on $\mathbb{R}^X$ in the general definition above, you will get the corresponding notion of convergence in that topology. For example, $f_n\to f$ in the product topology is just pointwise convergence: $$(\forall x\in X): f_n(x)\to f(x).$$ You can get this from the general definition by using the $\mathcal{N}_f$ for the product topology. For any other topology on $\mathbb{R}^X$, just apply the general definition with the appropriate notion of open set.

  • You might find this question helpful, where they consider the case $X = \mathbb{N}$.

Convergence in the product topology: A subbase for the product topology on $\mathbb{R}^X$ is given by $\{e_x^{-1}(U): x\in\mathbb{R}, \text{$U$ open in $\mathbb{R}$}\}$, where $e_x(f) = f(x)$ is the evaluation function. A sequence $(f_n)$ converges $f_n\to f$ in $\mathbb{R}^X$ iff $\forall x\in X$, $U$ open in $\mathbb{R}$, $f\in e_x^{-1}(U)$ implies that $f_n\in e_x^{-1}(U)$ for sufficiently large $n$.

This statement is equivalent to the pointwise convergence statement I wrote above. You can see that the $\forall x\in X$ appears in front for this topology.

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  • $\begingroup$ Thanks a lot for your thorough answer. Now, I am wondering on how the $x \in X$ enter in the picture for the pointwise convergence. To explain myself: my original (incorrect) definition was as such, because I felt I needed to put the $x \in X$ somewhere from the very beginning, because only later – thanks to the nature of the open sets taken into account – it would have entered more specifically in the definition of product topology. Now, I don't see how the $x \in X$ enters from the backdoor in the product topology, while not in the box or in the uniform topology. Could you clarify this? $\endgroup$
    – Kolmin
    May 21, 2015 at 8:04
  • $\begingroup$ It's a bit of a long explanation why $x\in X$ appears in different places for these different topologies. I'll add a little more explanation for the product topology case. $\endgroup$
    – b yen
    May 21, 2015 at 19:05
  • $\begingroup$ Thanks a lot again for the added explanation. I actually knew what you wrote, and – in some way – I felt it does not answer the question I wrote in the comment, which explicitly sounds like: If the only difference in the definition of box and product topology lies in the form of the open sets we consider, why do we have in one case (i.e. the product one) a $x \in X$ that pops up, while in the other not? But not, reading your answer, I am taken by an hypothesis. So (please, correct me if I am wrong), is what you mean that the subbase in the product topology gives us a natural reference... $\endgroup$
    – Kolmin
    May 22, 2015 at 6:40
  • $\begingroup$ ...to the $x \in X$ through the evaluation function, while there is no subbase of the box topology that actually refers to the $x \in X$ because that subbase of the product topology is not a subbase for the box topology, and there is no subbase for the box topology that looks conceptually like that one? (very long question) Again, thanks a lot. $\endgroup$
    – Kolmin
    May 22, 2015 at 6:42
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Certainly, the definition of convergence of sequences is the same in any topological space, and is as you stated.

What sequences converge to what, of course depends on the exact topology used. So it needs to be specified in context.

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  • $\begingroup$ First of all thanks a lot. The fact is that I am trying to see exactly the behoavior of convergence in those two topologies and in the uniform topology. It was (sort of) easy to compare the product and the uniform, where the uniform was defined metrically (i.e. $f_n \to f$ iff $\lim_{n \to \infty} \sup_{x \in X} \rho ( f_n (x) , f(x) )=0$), but - being self-taught - I was sort of lost when it came to the box topology. Btw, just wondering, is there a nhood counterpart of the definition of uniform convergence? $\endgroup$
    – Kolmin
    May 20, 2015 at 15:28
  • $\begingroup$ What is the topology of uniform convergence if not the one using the metric definition? $\endgroup$ May 20, 2015 at 16:19
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    $\begingroup$ I don't think the OP's formulation of the definition of convergence is correct. It should not have $\forall x\in X$, and $N(x)$ should be simply $N$. $\endgroup$ May 20, 2015 at 18:53
  • $\begingroup$ @AndreasBlass You're right of course. I overlooked that. $\endgroup$ May 20, 2015 at 20:36

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