Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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
1  
$[x;X] = C(X,Y)$ for any $x$. –  Henno Brandsma Jun 5 '13 at 3:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.