$C_p$ space topology

Question

If $X, Y$ are topological spaces then consider the set $C(X, Y) $ of continuous functions. Let $[x_1,...,x_n;O_1,...,O_n] =\{f\in C(X, Y) | f(x_i) \in O_i\in \tau (Y), i=1,...,n\}$. Then how to show that the collection $$\{U\subset C(X, Y) | \forall f\in U : \exists n\in \mathbb{N}\exists x_1,\dots,x_n \in X, O_1,\ldots, O_n \in \tau (Y):\\ f\in [x_1,...,x_n;O_1,...,O_n] \subset U \}$$ is a topology on $C(X, Y)$? I can see that $\emptyset $ is in there(vacuously) and $C(X, Y) $ is also in there since for any function $f\in C(X, Y) $ we can take any $x_1,...,x_n$ and $O_i=Y, i=1,..., n$. But I'm stuck with the other two.

Answer 1

The union is trivial: if $U_i, i \in I$ are all open and $f \in U=\bigcup_i U_i$, then for some $j \in I$, $f \in U_j$. By definition we then have $x_1,\ldots x_n$ in $X$ and $O_1,\ldots,O_n$ open in $Y$ such that $f \in [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \subseteq U_i \subseteq U$.

For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f \in U_1 \cap U_2$. Then $f \in U_1$ gives us $x_1,\ldots x_n$ in $X$ and $O_1,\ldots,O_n$ open in $Y$ such that $f \in [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \subseteq U_1$ and $f \in U_2$ gives us $x'_1,\ldots x'_m$ in $X$ and $O'_1,\ldots,O'_m$ open in $Y$ such that $f \in [x'_1,\ldots,x'_m; O'_1, O'_2,\ldots O'_m] \subseteq U_2$. But then

$$f \in [x_1,\ldots,x_n, x'_1,\ldots, x'_m; O_1,\ldots,O_n, O'_1,\ldots,O'_m] \subseteq \\ [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \cap [x'_1,\ldots,x'_m; O'_1, O'_2,\ldots O'_m] \subseteq U_1 \cap U_2$$

as required.

You can also note that basic open neighbourhood of $f$ (the $[x_1,\ldots,x_n; O_1, \ldots, O_n]$) is of the form $B \cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset $\{x_1, \ldots,x_n\}$ of "coordinates" from $X$ and finitely many $O_1, \ldots, O_n$ in those coordinates. So e.g. $f_n \to f$ in this topology iff $f_n(x) \to f(x)$ for all $x \in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.