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This is from Dudley´s book: Let $I:=[0,1]$ with the usual topology. Let $I^I$the set of all the functions from $I$ to $I$ with the product topology.

a) $I^I$ is separable. Hint: Consider function that are finite sums $\sum a_i1_{j(i)}$, where $a_i\in \mathbb{Q}$ and the $j(i)$ are intervals with the rational endpoints.

b) Show that $I^I$ has a subset which is not separable with the relative topology.

Sketch proof:

a) Let $A$ be the set of all the functions as described in the hint. We have to show that $A$ is indeed countable. There are countable many intervals with rational endpoints since each such interval can be identified with the pair $\langle a,b\rangle\in \mathbb{Q}\times \mathbb{Q}$ (for $a,b$ the endpoints) and the former is countable. Then any finite sequence of such intervals is countable. Since for such sequence of intervals we can always have a countable number of functions it follows that the family $A$ must be countable.

To conclude it will suffice to show that for any non-empty basic open set $U$ the intersection of $A$ and $U$ is non-empty. Let $U$ be a basic open set, then

$$U=\bigcap_{\{i\in F: F \text{ finite }\}}\text{pr}_i^{-1}(U_i)=\{f\in I^I:f(x)\in U_x\text{ and } x\in F \}$$

We take disjoint intervals with rational endpoints s.t. $x\in j(x)$ and also since $U_x\subset I$ for denseness of $\mathbb{Q}$ we choose $q_x\in {U_x}\cap \mathbb{Q}$. Hence $\sum_{x\in F} q_x1_{j(x)}$ is in $A\cap U$.

For b) we have to show that $\{0,1\}^I$ is not separable with respect to the relative topology...

The above argument is OK? , b) could someone give me a hint?

Thanks in advance :)

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  • $\begingroup$ I don't understand what you are saying in the paragraph preceding b), at all (but the countability argument is okay). About b), irrational numbers won't work. The general fact is, whenever you have a product of no more than $\mathfrak c$-many separable spaces, the product is separable (with much the same proof). (And the converse is true: a product of more than $\mathfrak c$-many nontrivial spaces is inseparable.) $\endgroup$ – tomasz Sep 6 '14 at 22:01
  • $\begingroup$ For the first part, see math.stackexchange.com/a/856096/4280, where I give essentially the same proof. $\endgroup$ – Henno Brandsma Sep 7 '14 at 8:41
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For the first part, I wrote this proof (written down differently) as this answer, so you could compare. It's essentially the same.

For (b), the subset $\{0,1\}^I$ is not going to work, as this is also separable; a generalisation of the argument from (a) will show that any product of at most $|[0,1]|$ (i.e. continuum many) separable spaces will be separable, so certainly $\{0,1\}^I$.

But consider the set $C = \{\chi_p: p \in [0,1]\}$, where $\chi_p: [0,1] \rightarrow [0,1]$ is defined by $\chi_p(x) = 0$ for $x \neq p$, $\chi_p(p) = 1$.

Then the open sets $U_p = (\pi_p)^{-1}[(\frac{1}{2},1]] = \{f \in [0,1]^{[0,1]}: f(p) \in (\frac{1}{2},1] \}$ is basic open in $[0,1]^{[0,1]}$ and is such that $U_p \cap C = \{\chi_p\}$, which shows that $C$ is discrete as a subspace, so cannot be separable, as it has size $|[0,1]|$. (Note also that adding the constantly $0$ function to $C$ gives us a copy of the one-point compactification of a discrete space of size $|[0,1]|$. This gives us a compact subspace that is non-separable.)

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