# Proof [0,1]^[0,1] is separable

This is a proof I made that $[0,1]^{[0,1]}$ is a separable space, please let me know if it has any flaws, its really long and I might have mistakes. I have a topology exam in a few days and this was an exam question, if this proof is satisfactory and understandable I am going to use this proof if the question is asked.

Notation: From now on Q=$ℚ∩[0,1]$

Since Q is countable an has cardinality $ℕ$, I can meke a bijective sucession $x_n$ in Q.

Then let $P_n$ = {$x_1,x_2,....,x_n$}, since $P_n$ is a finite set I can order it from smellest to bigges like so: $P_n$={${x_1}^n,....,{x_n}^n$}

Let $y=(y_1,...,y_n)$ $∈Q^n$ then I can define the function: $f_{n,y}$ $∈[0,1]^{[0,1]}$ as in: $f(x)= y_m + (y_{m+1}-y_m)*(x-{x_m}^n)/(x_{m+1}^n-{x_m}^n)$ if $x ∈ [{x_m}^n,x_{m+1}^n]$ and if $x>{x_n}^n$ then $f(x)= y_{n-1} + (y_n-y_{n-1})*(x-x_{n-1}^n)/(x_n^n-x_{n-1}^n)$

in other words the function connects the ''dots'' via lines, and when there is no more dots to connect it continues following the last line it was on.

Let $A_n$={$f_{n,y}$; $y ∈ Q^n$}, then for every natural ''n'' $A_n$ is a countable set

Let $A=∪A_n$ then $A$ is a countable union of countable sets, consequently it is countable. Now all there is left to proove is that A is dense in $[0,1]^{[0,1]}$.

let $g∈A^c$ and let $V$ be a neighborhood of $g$, without loss in generality I can assume $V$ is of the form $V=∩_{j∈J} π_j^{-1}(Uj)$ with $J$ a finite Real set.

Let n=$max${$j∈J$} then I can find a family of $z_1,....,z_n∈Q$ and $y_1,...,y_n∈Q$ such that $y_j ∈ U_j$ for every $j$ in $J$.

If $z_n=x_m$, then $P_m$ can be ordered in such a way that $z_i={x_m}_i^m$.

I can also define for every element in that family a point $a=(a_i)∈Q^n$ defined as $a_j$=$y_i$ if $j=m_i$ for some $i$ in $J$, and $a_j=0$ if not.

Then I can select $z_1,....,z_n$ and $y_1,...y_n$ with $z_i$ is close enough to $i$ such that the line that connects $(z_i,f_{n,a}(z_i))$ and $(y_i,{x_m}_{i+1}^m)$ passes through $U_i$. (This is because $y_i∈ U_i$ and a line is a continuous function) (Note: remember $z_i= {x_m}_i^m$

Thus by definition $f_{n,a}$ is in the intersection of $A$ an $V$. Since I can do this for every $g∈A^c$ then A is dense in $[0,1]^{[0,1]}$.

I appreciate the feedback...Thanks in advance

• The idea is fine. You can also use a countable base$\mathcal{B}$ for $[0,1]$ and consider all finite pairwise disjoint subsets from $\mathcal{B}$ and for each such set $\mathcal{B'} = \{B_1, \ldots B_n\}$ of size $n$ and an $n$-tuple $(q_1,\ldots, q_n)$ from $\mathbb{Q} \cap [0,1]$, the function that constant with value $q_n$ on $B_n$, and $0$ elsewhere. No messing about with lines; the functions need not be continuous just dense. Commented Mar 3, 2017 at 19:33
• Ok, because any second-countable space is separable, the idea is finding a countable base. But maybe finding a countable base might not be that hard. I will try to do it that way, since its simpler. Commented Mar 3, 2017 at 22:00
• Though Im pretty sure [0,1]^[0,1] is not second-countable. If you want, I think I have a proof of that. Commented Mar 3, 2017 at 22:37
• $X = [0,1]^{[0,1]}$ is not even first countable, let alone second countable. This holds, as when we would have a countable local $\{B_n: n \in \mathbb{N}\}$ base for $f \in X$, we could refine each of the $B_n$ with a basic open set depending on finitely many points $F_n \subset [0,1]$ and finitely many open sets $O(n)(x), x \in F_n$. Then $\cup_n F_n$ is countable, so pick $p \in [0,1]\setminus \cup_n F_n$. Commented Mar 4, 2017 at 6:01
• You then check that a basic neighbourhood depending on this $p$ and some non-trivial open set $O$ containing $f(p)$ cannot contain any of the $B_n$, as these leave the values of points outside of the $F_n$ completely free. So the $B_n$ cannot be a local base at $f$. Commented Mar 4, 2017 at 6:01

The proof idea is fine, though I have some thoughts on the proving of being dense.

A basic open set of $X = [0,1]^{[0,1]}$ has the form $B(x_1, \ldots,x_n ;O_1, \ldots O_n) = \{f \in X: \forall i: f(x_i) \in O_i\}$, determined by finitely many points of $[0,1]$ and finitely many open sets in which their images must lie. (The $x_n$ are your $J$ above, the $O_i$ your $U_i$.)

Your candidate dense functions are piecewise linear functions determined by finitely many pairs of rationals. So we cannot expect the $x_i$ to be rational, so you have to do some extra work to see that there is some function from your set that does map $x_i$ into $O_i$, using some argument that's not terribly well written up (IMHO), but seems to use the continuity of your piecewise linear functions in some essential way. You would probably need to pick 2 nearby rationals on both sides of the $x_i$ and mapping these into rationals in $O_i$; so this can be made to work, I believe. It's notationally a bit awkward, as you notice (special cases at the boundary etc.)

It's much easier I think (and most text books would agree I think) to use a countable base $\mathcal{B}$ of $[0,1]$ (e.g. the intervals with endpoints in $Q$) and consider the set of pairwise disjoint finite families

$$\mathcal{F} = \{(B_1,\ldots,B_n)\in \mathcal{B}^n: n \in \mathbb{N}: \forall i \neq j: B_i \cap B_j = \emptyset\}$$

from this base $\mathcal{B}$. As the base is countable, so is this set (this needs a little argument, maybe, about countable unions of countable sets, I don't know how much you're allowed to assume in that), and for each pair $\underline{B} = (B_1, \dots, B_n) \in \mathcal{F}$ and $\underline{q}= (q_1, \ldots q_n) \in Q^n$ we define $f = f_{\underline{B}, \underline{q}}$ from $[0,1]$ to $[0,1]$ by $f(x) = q_i$ for $x \in B_i$, $f(x) = 0$ for $x \notin \cup_i B_i$. This is well-defined by disjointness. The set of all these $f$ is still countable as $Q^n$ is.

And if $B(x_1, \ldots,x_n ;O_1, \ldots O_n)$ is a basic open set we find pairwise disjoint basic open sets $B_i$ with $x_i \in B_i$, say refining disjoint intervals, and $q_i \in O_i$ by density of the rationals.

Then by definition $f_{(B_1,\ldots, B_n), (q_1, \ldots q_n)} \in B(x_1, \ldots,x_n ;O_1, \ldots O_n)$. The construction immediately implies the density, which simplifies things. The notational complications are limited in this case. The essential idea remains the same though.