Post History
#2: Post edited
by
clemens
·
2026-04-27T23:43:05Z (5 months ago)
*Disclaimer*: this is not a complete answer. Indeed, the OP has likely thought of some or all of the possibilities mentioned. I answer in the hope that it will be helpful to other users and perhaps provoke them to write a more complete answer.- ---
- In the special case where all the factors in a product of spaces are the same, there are two topologies I know of that are commonly used:
- 1. The *topology of uniform convergence*, wherein a sequence/filter of functions $f_n$ converges to $f$ iff it converges uniformly. This is the natural topology for sequences or functions in whose uniform convergence we are interested; the box topology, by contrast is the topology of pointwise convergence.
- 2. The *compact-open topology*, which is generally used for spaces of continuous functions $\mathcal{C}(\text{Dom}, \text{Cod})$. It is weaker than the uniform topology and I think *strictly weaker* iff the domain $\text{Dom}$ of the functions is not compact. The compact-open topology captures the intuition that we ought to be able not only to approximate *particular values* of a continuous function (as we can in the product topology) but be able to obtain ever more precise *bounds* of its value over "reasonable" (namely compact) regions.
- In the case where the factors are *not* necessarily the same, the topology of uniform convergence seems to be a pretty natural way of taking the product of metric topologies, though I don't know off the top of my head where this concept has been used.
- This topology can also be generalized to a product of arbitrary subsets of a uniform topological space: just take as one's basis the product of open sets taken from some uniform cover. This reduces to the topology of uniform convergence if the uniform topological space is a metric space.
- As for the OP's idea of "demand[ing] that at most countably many factors are not the full space", I admit I couldn't figure out much to do with it. I thought that maybe one could prove something analogous to the Alexander subbase lemma and then derive a theorem that the product of Lindelöf spaces, under this topology, is Lindelöf, but I'm pretty sure that won't work.
- *Disclaimer*: this is not a complete answer. Indeed, the OP has likely thought of some or all of the possibilities mentioned. I answer in the hope that it will be helpful to other users and perhaps even spur them on to think of a more complete answer.
- ---
- In the special case where all the factors in a product of spaces are the same, there are two topologies I know of that are commonly used:
- 1. The *topology of uniform convergence*, wherein a sequence/filter of functions $f_n$ converges to $f$ iff it converges uniformly. This is the natural topology for sequences or functions in whose uniform convergence we are interested; the box topology, by contrast is the topology of pointwise convergence.
- 2. The *compact-open topology*, which is generally used for spaces of continuous functions $\mathcal{C}(\text{Dom}, \text{Cod})$. It is weaker than the uniform topology and I think *strictly weaker* iff the domain $\text{Dom}$ of the functions is not compact. The compact-open topology captures the intuition that we ought to be able not only to approximate *particular values* of a continuous function (as we can in the product topology) but be able to obtain ever more precise *bounds* of its value over "reasonable" (namely compact) regions.
- In the case where the factors are *not* necessarily the same, the topology of uniform convergence seems to be a pretty natural way of taking the product of metric topologies, though I don't know off the top of my head where this concept has been used.
- This topology can also be generalized to a product of arbitrary subsets of a uniform topological space: just take as one's basis the product of open sets taken from some uniform cover. This reduces to the topology of uniform convergence if the uniform topological space is a metric space.
- As for the OP's idea of "demand[ing] that at most countably many factors are not the full space", I admit I couldn't figure out much to do with it. I thought that maybe one could prove something analogous to the Alexander subbase lemma and then derive a theorem that the product of Lindelöf spaces, under this topology, is Lindelöf, but I'm pretty sure that won't work.
#1: Initial revision
by
clemens
·
2026-04-27T23:42:40Z (5 months ago)
*Disclaimer*: this is not a complete answer. Indeed, the OP has likely thought of some or all of the possibilities mentioned. I answer in the hope that it will be helpful to other users and perhaps provoke them to write a more complete answer.
---
In the special case where all the factors in a product of spaces are the same, there are two topologies I know of that are commonly used:
1. The *topology of uniform convergence*, wherein a sequence/filter of functions $f_n$ converges to $f$ iff it converges uniformly. This is the natural topology for sequences or functions in whose uniform convergence we are interested; the box topology, by contrast is the topology of pointwise convergence.
2. The *compact-open topology*, which is generally used for spaces of continuous functions $\mathcal{C}(\text{Dom}, \text{Cod})$. It is weaker than the uniform topology and I think *strictly weaker* iff the domain $\text{Dom}$ of the functions is not compact. The compact-open topology captures the intuition that we ought to be able not only to approximate *particular values* of a continuous function (as we can in the product topology) but be able to obtain ever more precise *bounds* of its value over "reasonable" (namely compact) regions.
In the case where the factors are *not* necessarily the same, the topology of uniform convergence seems to be a pretty natural way of taking the product of metric topologies, though I don't know off the top of my head where this concept has been used.
This topology can also be generalized to a product of arbitrary subsets of a uniform topological space: just take as one's basis the product of open sets taken from some uniform cover. This reduces to the topology of uniform convergence if the uniform topological space is a metric space.
As for the OP's idea of "demand[ing] that at most countably many factors are not the full space", I admit I couldn't figure out much to do with it. I thought that maybe one could prove something analogous to the Alexander subbase lemma and then derive a theorem that the product of Lindelöf spaces, under this topology, is Lindelöf, but I'm pretty sure that won't work.