Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable. Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.
According to Munkres Topology:
Definition: A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\beta$.
A space that has a countable basis at each of its points is said to be first countable.
I can also proceed indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) dt$ where $f$ is measurable. But I don't know how to prove that there exists such a function. Any help please?
 A: Simply take $f$ to be the function that is constant $0$ (but any function will do).
Suppose, for a contradiction, that $f$ has a countable local base $\{U_n\mid n \in \Bbb N\}$ in $\{0,1\}^{\Bbb R}$ (in the product topology).
For every $n$ we can find a basic product open subset of the form $B_n:=\prod_{r \in \Bbb R} O_r$, where each $O_r$ is open in $\{0,1\}$, and such that there is a finite subset $F_n \subseteq \Bbb R$ such that for all $r \notin F_n$ we know that $O_r=\{0,1\}$ and such that $$f \in B_n=\prod_{r \in \Bbb R} O_r \subseteq U_n\tag{1}$$
We thus know also that for $r \in F_n$ we have $O_r =\{0\}$ (the only non-trivial (open) subset that contains $0$ in $\{0,1\}$), although we won't use it. We could also have written the basic open subset as $\bigcap_{n \in F_r} \pi_r^{-1}[O_r]$, if that's more familiar to you.
So for each $n$ we have such a basic open subset that depends essentially only on a finite subset $F_n$ of $\Bbb R$ that sits inside $U_n$. Now, $\bigcup_{n \in \Bbb N} F_n$ is at most countable and so I can pick some real number $r_0 \notin \bigcup_{n \in \Bbb N} F_n$. Then define $O= \pi_{r_0}^{-1}[\{0\}] = \{g \in \{0,1\}^{\Bbb R}\mid g(r_0)=0\}$, which is an open subset of the product (it's basic open even) that contains $f$, the constantly $0$ function.
Because the $\{U_n\mid n \in \Bbb N\}$ form a local base at $f$ by assumption, we have that form some $m \in \Bbb N$:
$$U_m \subseteq O\tag{2}$$
But now define $g: \Bbb R \to \{0,1\}$ by $g(r_0)=1$ and $g(r)=0$ for all $r \neq r_0$.
Then in particular $g(r)$ is $0$ for all $r \in F_m$ (as $r_0 \notin F_m$ by how it was chosen!) and so $$g \in B_m$$ and so $$g \in U_m \subseteq O$$ by $(1)$ and $(2)$. But this contradicts that $g \notin O$ (as $g(r_0)= 1 \neq 0$)!  This contradiction shows that our initial assumption that $f$ has a countable local base could not have been not correct. QED.
Note finally that we can do this proof at any point of $\{0,1\}^{\Bbb R}$, so nowhere there is a local countable base and we can even further generalise this (if you know simple cardinal arithmetic) that there cannot even be a local base of size $< |\Bbb R|$, which is a stronger fact in general.
A: You can show that there is no countable basis at $0$ in the following way.
First, notice that if there is a countable basis, this implies that any basis has a countable subset which is also a basis, to it is enough to find a basis at $0$ whose countable subsets are all not bases.
Consider, then, for each finite $A\subseteq \mathbf R$ the set $U_A:=\{x\in X\mid (\forall a\in A)\, x(a)=0 \}$. This is clearly a basis at $0$.
The conclusion follows from the observation that if $\{A_n\mid n\in \mathbf N\}$ is a countable family of finite subsets of $\mathbf R$, then $\mathbf R\setminus \bigcup_n A_n$ is nonempty, so for any $r\in \mathbf R\setminus \bigcup_n A_n$, the set $U_{\{r\}}$ is a neighbourhood of $0$ which does not contain $\bigcap_n U_{A_n}$, so it certainly does not contain any of the sets $U_{A_n}$.
Hence, $\{U_{A_n}\mid n\}$ is not a basis.
A: You can also argue indirectly as follows:
$\{0,1\}^\mathbb R$ is compact because it is a product of compacts, by Tychonoff's theorem.  However, it is not sequentially compact (see this question). Since first countable, compact spaces are sequentially compact (see this question), it follows that $\{0,1\}^\mathbb R$ is not first countable.
