pointwise convergence is equivalent to the convergence under the topology of pointwise convergence In the note 
http://math.bard.edu/belk/math351/FunctionSpaces.pdf
thereom 2 says that $f_n \rightarrow f$ under the product topology/topology of pointwise convergence if and only if the function $f_n$ converges pointwise to $f$ in the usual sense.
To the prove the forward direction, I guess more specific argument is:
for each $x\in X$, let $U$ be a neighbourhood of $f(x) \in Y$, then $S(x,U)$ is a neighbourhood of $f$ under the topology of pointwise convergence. Then, $f_n \rightarrow f$ under the product topology/topology of pointwise convergence means that for any $x\in X$ there exists $N_x$ (since this depends on the choice of $x$) such that for all $n \geq N_x$, $f_n \in S(x,U)$ for some $U$. Next we can take the maximal $N$ of $N_x$ over $x\in X$, and say that $f_n(x) \in U$ for all $n \geq N$.
This argument is correct, right?
So I was wondering what happens if $X$ is infinite? could we still take the maximal $N$ of $N_x$ over $x\in X$? is it possible that $N_x$ tends to infinite with some sequence of $x$?
 A: What you have is not quite right. The argument does not require looking at more than one coordinate at a time.
We’re assuming that $\langle f_n:n\in\Bbb N\rangle\to f$ in the product topology on $Y^X$, and we want to show that for each $x\in X$, $\langle f_n(x):n\in\Bbb N\rangle\to f(x)$. This means that for each $x\in X$ we want to show that if $U$ is any open nbhd of $f(x)$ in $Y$, there is an $m_x\in\Bbb N$ such that $f_n(x)\in U$ for each $n\ge m_x$.
Fix $x\in X$ arbitrarily, and let $U$ be an open nbhd of $f(x)$ in $Y$. Then
$$S(x,U)=\left\{g\in Y^X:g(x)\in U\right\}$$
is by definition an open set in $Y^X$. Since $\langle f_n:n\in\Bbb N\rangle\to f$ in $Y^X$, there is an $m\in\Bbb N$ such that $f_n\in S(x,U)$ for each $n\ge m$. But by the definition of $S(x,U)$ this just means that $f_n(x)\in U$ for each $n\ge m$, so we can take $m_x=m$. This shows that for each $x\in X$ and open nbhd $U$ of $f(x)$ in $Y$ there is indeed an $m_x\in\Bbb N$ such that $f_n(x)\in U$ for each $n\ge m_x$, and hence by definition $\langle f_n(x):n\in\Bbb N\rangle\to f(x)$.
Since $x\in X$ was arbitrary, this shows that $\langle f_n:n\in\Bbb N\rangle$ converges pointwise to $f$ in the usual sense.
