5

The standard topology on $\mathbb{R}$, the set of real numbers, has as a basis all open intervals $(a,b)$, where $a$ and $b$ are real numbers such that $a < b$.

Let $J$ be an arbitrary (finite, countable, or uncountable) index set, and let $\mathbb{R}^J$ denote the set of all $J$-tuples $\mathbf{x} = \left( x_\alpha \right)_{\alpha\in J}$ of real numbers (i.e. the set of all functions $\mathbf{x} \colon J \longrightarrow \mathbb{R}$).

Then the box topology on $\mathbb{R}^J$ by definition is the topology having as a basis all sets of the form $\prod_{\alpha \in J} U_\alpha$, where $U_\alpha$ is open in $\mathbb{R}$ with the standard topology for each $\alpha \in J$.

The product topology on $\mathbb{R}^J$ has as a basis all sets of the form $\prod_{\alpha \in J} U_\alpha$, where $U_\alpha$ is open in $\mathbb{R}$ with the standard topology for each $\alpha \in J$ and the sets $U_\alpha$ are distinct from $\mathbb{R}$ for at most finitely many indices $\alpha \in J$.

Finally, the uniform topology on $\mathbb{R}$ is the topology induced by the uniform metric $\bar{\rho}$ on $\mathbb{R}^J$, which is defined as follows: $$ \bar{\rho}( \mathbf{x} , \mathbf{y} ) := \sup \left\{ \, \min \left\{ \left\lvert x_\alpha - y_\alpha \right\rvert, 1 \right\} \colon \alpha \in J \, \right\} $$ for all $\mathbf{x} := \left(x_\alpha\right)_{\alpha \in J}$, $\mathbf{y} := \left(y_\alpha\right)_{\alpha \in J}$ in $\mathbb{R}^J$.

Now Munkres on pp 124--125 has given a proof that the uniform topology on $\mathbb{R}^J$ is finer than the product topology but coarser than the box topology, but toward the end of the proof he states that when the index set $J$ is infinite, then these three topologies are different.

How?

Can we explicitly exhibit a set that is in the box topology but is not in the uniform topology?

And, can we then explicitly exhibit a set that is in the uniform topology but is not in the product topology?

Just now, I don't seem to be able to think of such examples.

I would of course prefer this elementary, direct approach rather than taking any convoluted, indirect route.

Saaqib Mahmood
  • 26,762
  • 1
    See, for example, Exercise 4 of Section 20. The approach here, then, is to show that certain sequences converge in some of the topologies but not all of them. This is "indirect", but I would not say it is convoluted. For more "direct", see Exercise 6 of the same section, which provides a set open in the box topology but not the uniform topology (and hence not in the product topology). There's probably another exercise giving an explicit example of something open in the product but not uniform, but I don't remember which. – Hayden Mar 27 '15 at 17:51

1 Answers1

1

if the index is infinite then the box topology has open sets with have arbitrarily small projections. e.g. for a countable index set, consider the set with projections:

$$ U_j = B(0,\frac1{n}) $$ this is open in the box topology, but contains no open ball of the uniform topology.

David Holden
  • 18,040
  • I guess you mean that the set $U \colon=(-1,1)\times(-\frac{1}{2}, \frac{1}{2})\times(-\frac{1}{3}, \frac{1}{3})\times\cdots$ is open in the box topology on $\mathbb{R}^\omega$. Fair enough. And, the point $(0,0,0,\ldots) \in U$, but how do we show that this set is not open in the uniform topology? – Saaqib Mahmood Mar 27 '15 at 18:25
  • to define a uniform open set inside $U$ you would have to find an $\epsilon \gt 0$ which satisfies $\forall n \in \mathbb{N} \epsilon \lt \frac1{n} $ – David Holden Mar 27 '15 at 21:24