# Topology of pointwise convergence and point-open topology are equivalent

Definition VII.1. In Nagata's Modern General Topology construct Topology of pointwise convergence as follows:

Given $x_1,\cdots,x_n\in X$ and $O_1,\cdots, O_n\in \mathcal{T}_Y$, let $$[x_1,\cdots,x_n;O_1,\cdots, O_n]=\{f\in C(X,Y): f(x_i)\in O_i\}\;.$$ Then the family $$PC(X, Y)=\{U\subset C(X, Y) : \text{for any } f \in U \text{ we have }f \in [x_1,\cdots,x_n;O_1,\cdots, O_n]\subset  U \text{ for some } n\in \Bbb N, x_1,\cdots,x_n\in X \text{ and } O_1,\cdots, O_n\in \mathcal{T}_Y\}$$ is called the topology of pointwise convergence on the set $C(X, Y)$.

1) $PC(X,Y)$ is a topology or a base for a topology on $C(X,Y)$?

The other references are defined point-open topology on $C(X)$ as follows:

Let $\Bbb F(X)$ denote the set of all finite subset of $X$. for $A\in \Bbb F(X)$ and an open set $V$ of $\Bbb R$, define $[A,V]=\{f\in C(X): f(A)\subseteq V\}$. The collection $\{[A,V]:A\in\Bbb F(X), V\text{ open in }\Bbb R\}$ forms a subbase for the point-open topology on $C(X)$.

2) Do these two definitions are equivalent?

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1) $PC(X,Y)$ is the topology itself. The sets $[x_1,\ldots,x_n;O_1,\ldots,O_n]$ form a base, by definition.

These sets are closed under intersections:

$$[x_1,\ldots,x_n;O_1,\ldots,O_n] \cap [x'_1,\ldots,x'_m;O'_1,\ldots,O'_m] = [x_1,\ldots,x_n,x'_1,\dots,x'_m; O_1,\ldots,O_n,O'_1,\ldots,O'_m]$$

and they cover $C(X,Y)$, clearly, so they form the base for some topology.

Note that this is just the subspace topology $C(X,Y)$ as a subspace of the product topology on $Y^X = \prod_{x \in X} Y$, all functions (not necessarily continuous) from $X$ to $Y$. This holds because these sets are basically just product sets that only depend on finitely many coordinates (the $x_i$) and have no limitations on the other coordinates.

2) Yes, these topologies are equivalent. The sets $[A;V]$ also form a base for $PC(X,Y)$ from 1. They are certainly open: if $A = \{x_1,\ldots,x_n\}$ then $[A;V] = [x_1,\ldots,x_n;V,\ldots,V]$. And $$[x_1,\ldots,x_n;O_1,\ldots,O_n] = \cap_{i=1}^n [\{x_i\}; O_i]$$ so the other basic elements are open in this topology too. So the basic sets are "mutually open" so the topologies coincide.

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How we can show that the sets $[x_1,\ldots,x_n;O_1,\ldots,O_n]$ cover $C(X,Y)$? $X,Y$ must be tychonoff? – TXC Jun 5 '13 at 2:27
$[x;X] = C(X,Y)$ for any $x$. – Henno Brandsma Jun 5 '13 at 3:41