Open and closed sets in the space of real valued functions on $[0,1]$ Consider the space $F$ of real valued functions on $[0,1]$ given the product topology. Then can we say that the subspace of continuous functions is closed in $F$? If not, then what about the space of integer-valued, bounded or unbounded functions?
This question is perturbing. What is a basis of the space $F$? Is it a product of open sets of $\mathbb{R}$? Any hints? Thanks beforehand. 
 A: The product topology is generated by a base where only in finitely many coordinates (values of the domain) we can restrict the values to some $\varepsilon$-range. The rest is left free. So a typical basic neighbourhood of a function $f: X \to \Bbb R$ (I’m working with a general domain $X$ for generality) looks like:
$$U(f; F; r)= \{g:X \to \Bbb R\mid \forall x \in F: |f(x)-g(x)| < r\}$$
where $F\subseteq X$ is finite and $r>0$ is in $\Bbb R$.
This means that the continuous functions are dense in all functions on $X$ when $X$ is completely regular and $T_1$: we can always  find a continuous function with predetermined values on a finite set of domain points. This holds in particular in $X=[0,1]$ and all metric space domains.
Unbounded functions are also dense in all functions, and so are bounded ones. We only have finitely many constraints to fulfill to intersect a basic open set. Integer valued (general) functions are closed in all functions, only one witnessing point is enough for a function not to integer-valued and a basic open set can take advantage of this.
A: Answer for the first part: If you take a sequence of continuous functions on $[0,1]$ converging point-wise to a discontinuous function (say $(x^{n})$) then we get a sequence in $F$ converging to an element not in $F$. So $F$ is not closed.
A: To answer the second part: the basis of F in the product topology is given by products of open sets of $\mathbb{R}$ of the form $\prod_{x \in [0,1]} U_x$  where $U_x = \mathbb{R}$ for all but finitely many indices, and is an arbitrary open set for those finite indices. This is because the product topology is the initial topology with respect to the projection maps $\pi_x: F \to \mathbb{R}, f\ \mapsto f(x), x \in [0,1]$. This yields a sub-basis given by $\{\pi_x^{-1} (U_x) : x \in [0,1], U_x \subseteq \mathbb{R} \text{ open } \}$ which equals $\{f \in F: f(x) \in U_x, x \in [0,1], U_x \subseteq \mathbb{R} \text{ open } \}$. Finite intersections of sets in this sub-basis can be seen to be equal to the product defined at the start of this paragraph (since the intersections can only be finite, we only get to specify a finite number of indices for basis sets in the product).
