Definition of convergence in the product and in the box topology

Question

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.

Answer 1

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 $X = \mathbb{N}$.

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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.

Answer 2

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.