Do pointwise topology and product topology coincide on $[0,1]^{[0,1]}$? The pointwise topology is generated by the sub-basis given by the collection $$S(x,U)=\{f:[0,1]\to [0,1]\mid f(x)\in U\},$$
where $x\in [0,1]$ and $U$ is open in $[0,1]$. We also have that $$[0,1]^{[0,1]}=\prod_{x\in [0,1]}[0,1].$$
On $\prod_{x\in [0,1]}[0,1]$ we have the sub-basis given by elements of the form $$\prod_{x\in [0,1]}V_x,$$
where $V_x$ open in $[0,1]$ and $V_x\neq [0,1]$ for finitely but many $x$. 
My question Does both topology coincides ? It looks to coincides since at the end $\{f:[0,1]\to [0,1]\mid f(x)\in U\}$ looks to be $\prod_{y\in [0,1]}V_y$ where $V_y=[0,1]$ if $y\neq x$ and $V_x=U$, but I'm really not sure about that.
 A: Actually a much more general case is true. Suppose $X$ is an arbitrary set (not even a topological space) and $Y$ is a topological space.
Then the pointwise topology on $Y^X$ is identical to the product topology on $\prod_{X}Y$.
And as you noticed, one of the directions in demonstrating this is noticing that the subbasis element $S(x, U)$ of $Y^X$ (for $x \in X$ and open $U \subseteq Y$) is exactly the basis element $\prod_{s \in X}U_s$ of $\prod_{X}Y$ (where $U_x = U$ and $U_s = Y$ for $s \neq x$) :


*

*Because if $f \in \prod_{s \in X}U_s$, then it is a function $f : X \to Y$ such that $f(s) \in U_s$. In particular $f(x) \in U_x = U$. So $f \in S(x, U)$.

*And if $f \in S(x, U)$, then it is a function $f : X \to Y$ such that $f(x) \in U = U_x$. Also, if $s \neq x$, then $f(s) \in Y = U_s$. Thus, $f \in \prod_{s \in X}U_s$.


Going the other direction, any basis element $\prod_{s \in X}U_s$ of $\prod_X Y$, where $U_s = Y$ for all but finitely many open sets $U_{s_1}, \ldots, U_{s_n} \subseteq Y$, can be generated by subbasis elements of $Y^X$: $$
\prod_{s \in X}U_s = S(x, U_{s_1}) \cap \cdots \cap S(x, U_{s_n})
$$ Note: as there are only finitely many intersections, the result remains validly open in $Y^X$.
Thus the topologies generated by the respective subbases are the same. And of course your question is about the special case when $X = [0, 1]$ and $Y = [0, 1]$ with the usual topology.
A: These topologies do agree.
The key is to note that $S(x,U) = \pi_x^{-1}(U)$ where $\pi_x: [0,1]^{[0,1]} \to [0,1]$ defined by $f\in [0,1]^{[0,1]}\mapsto f(x)$ is the projection map, and the preimages of the projections are usually taken to be a subbasis of the product topology. Finite intersections of the above mentioned preimages are a basis of the product topology as mentioned in the question.
So the pointwise and the product topology as defined in the question coincide.
