Explanation of uniform topology theorem in Munkres

[mathmonkey]

Hi all,

I'm looking for some help in understanding one of the theorems stated in section 20 of Munkres. The theorem is as follows:

The uniform topology on $\mathbb{R}^J$ (where $J$ is some arbitrary index set) is finer than the product topology and coarser than the box topology; these three topologies are all different if $J$ is infinite.

This theorem seems to break down into two cases: $J$ finite, or $J$ infinite. In the case that $J$ is finite, aren't the box and product topologies equivalent? Hence it seems like the first sentence of the theorem can be strengthened to say all three topologies are equivalent?

For the second case, if $J$ is infinite, I thought that the box topology is finer than the product topology, since the product topology has the restriction that for each basis element, only finitely many of each of the $\mathbb{R}_i$ are open such that they are not equal to $\mathbb{R}_i$ itself, whereas the box topology does not have this restriction. So I'm not sure I understand why these topologies are different in the case that $J$ is infinite? All the statements in this theorem seem to contradict what I understood from the previous chapter on product topologies.

Any help explaining the theorem is appreciated. Thanks!

 

[micromass]

Hi all,

I'm looking for some help in understanding one of the theorems stated in section 20 of Munkres. The theorem is as follows:

The uniform topology on $\mathbb{R}^J$ (where $J$ is some arbitrary index set) is finer than the product topology and coarser than the box topology; these three topologies are all different if $J$ is infinite.

This theorem seems to break down into two cases: $J$ finite, or $J$ infinite. In the case that $J$ is finite, aren't the box and product topologies equivalent? Hence it seems like the first sentence of the theorem can be strengthened to say all three topologies are equivalent?

Correct.

For the second case, if $J$ is infinite, I thought that the box topology is finer than the product topology, since the product topology has the restriction that for each basis element, only finitely many of each of the $\mathbb{R}_i$ are open such that they are not equal to $\mathbb{R}_i$ itself, whereas the box topology does not have this restriction.

Correct.

So what we're claiming is that the box topology is finer than the uniform topology, and that the uniform topology is finer than the product topology.

 

[mathmonkey]

Oh...I could've sworn I read somewhere earlier in the text that two topologies are defined to be different if neither is finer or coarser than the other. So I suppose "different" in this case just means they are not equal?

I guess I just bashed my head over the table for an hour over misunderstanding of wording then . At least its good to see my previous understanding of the topic wasn't wrong though.

 

[Bacle2]

Just curious, I don't have Munkres --nor his book :) --with me. Is the uniform topology

the topology of uniform convergence in function spaces?