Why is the infinite Cartesian product of [0,1] with itself not countably based?

[Extropy]

A question about bases

I have a question as to why a particular space does not have a countable base. The space is the infinite Cartesian product of the interval [0,1] with the usual topology, with itself.

 

[HallsofIvy]

?? The interval [0, 1], with the usual topology, itself, does not have a countable base. why would you expect an infinite Cartesian product of such things to have a countable base?

 

[CompuChip]

Now you've got me confused, sorry.
It's been a while, but isn't
$$\{ (q - 1/n, q + 1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N} \} \cap [0, 1]$$
one?

(which still wouldn't work, as for any finite product this is still countable, but for an infinite product it's not).

Probably you should / could / would have to prove that the space is not second countable.

 

[ZioX]

R^n with the usual topology is second countable.

 

[CompuChip]

For all $n \in \mathbb{N}$, yes. But what about $n \to \infty$?

 

[morphism]

For all $n \in \mathbb{N}$, yes. But what about $n \to \infty$?

It's still second countable: a countable basis is the collection of products of the form $\prod_n X_n$ where X_n is an interval with rational endpoints for finitely many n and X_n = R.

And [0,1] does have a countable base for its usual topology -- after all, it's a separable metric space.

Now as to why [0,1]x[0,1]x... isn't second countable, well, how big of a product are we talking about? If it's a countable product then it is second countable, but if it's an uncountable product then it's not even first countable.

 

[Extropy]

Now as to why [0,1]x[0,1]x... isn't second countable, well, how big of a product are we talking about? If it's a countable product then it is second countable, but if it's an uncountable product then it's not even first countable.

Well, as big as a product to make it uncountable, if there is such a product. Could you explain how the uncountable product of [0,1] with itself is not countable?

 

[morphism]

(You mean second countable.)

Let's take the product of [0,1] with itself [0,1] times, i.e. [0,1]^[0,1], as prototypical example. This is just another way to talk about the set of functions from [0,1] into [0,1]. A nbhd of a function f in [0,1]^[0,1] (in the product topology) is given by a product of open sets in [0,1] all but finitely many of which are [0,1]. We can picture these nbhds as wiggle room for f, where we can restrict the wiggling at only finitely many points in [0,1].

Now suppose {B_i} is a countable local base at f, and write B_i = $\prod_{a \in [0,1]} U_{i,a}$, where all but finitely many of the U_{i,a}'s are [0,1]. Now because [0,1] is uncountable, we know there exists an a* in [0,1] such that U_{i,a*} = [0,1] for all i. But this means we can't restrict the wiggle space of f at a*, which is a contradiction. More formally, the nbhd $\prod_{a \in [0,1]} G_{a}$, where G_{a*} is strictly smaller than [0,1] (but big enough to contain f(a*)) and G_a = [0,1] for all other a, of f doesn't contain any B_i. So {B_i} can't possibly be a local base at f.

So [0,1]^[0,1] isn't first countable, and consequently not second countable.

This might be too much to swallow if you're not comfortable with the product topology and viewing Cartesian products as sets of functions, but if you are, try putting my words above into picture and the details will write themselves.