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.

In general, let $X$ be a set, $\mathcal N:X \to \wp(\wp(X))$ assign neighbourhood system to every point, $B$ be a filter basis on $\wp(X)$, then we say $B$ converges to $x$, or $B \to x$ iff
$$\forall U \in \mathcal{N}_x \exists F \in B(F \subseteq U)$$ in which $x \in X$, $\mathcal{N}_x$ is the neighbourhood system of $x$.

However, the pointwise convergence of function is different, let $D$ be a filter basis on $\wp(Y^X)$, $D$ pointwise converges to $f$ iff $Dx \to f(x)$ for all $x \in X$ in witch $f\in Y^X$, $Dx=\{Gx|G \in D\}$, in which $Gx=\{g(x)|g \in G\}$.

My question is can we take the pointwise convergence of function also a convergence of point? Since functions can also be seen as points.

My idea is find a definition of $\mathcal N_f$ such that make these two formula equivalent:

(1)$\forall V \in \mathcal N_f\exists G \in D(G \subseteq V)$

(2)$\forall x \in X\forall U \in \mathcal N_{f(x)}\exists G \in D(Gx \subseteq U)$

share|improve this question
    
You mean $Y^X$ instead of $X^Y$, right? –  martini Jul 6 '12 at 9:09
2  
Wikipedia: Pointwise convergence is the as convergence in the product topology on the space $Y^X$. –  Martin Sleziak Jul 6 '12 at 9:23
    
@martini Yes, thanks. –  Popopo Jul 8 '12 at 15:19
add comment

1 Answer 1

up vote 1 down vote accepted

I'm not quite sure, what you want for a general "nbhd system" to hold, but the following seems to work: Let's try to follow the idea of the product topology, that is we define $\mathcal N_f$ to be the filter generated by \[ \left\{ \prod_{x\in X} U_x \biggm| U_x \in \mathcal N_{f(x)}, U_x \ne Y \text{ only finitely often} \right\} \]

Now suppose (1) holds. Let $x \in X$, $U \in \mathcal N_{f(x)}$, define $V := Y^{X \setminus\{x\}} \times U$, then $V \in \mathcal N_f$, by (1) there is a $G \in D$ with $G \subseteq V$, so $Gx \subseteq Vx = U$. So (2) holds.

Now suppose (2) holds, let $V \in \mathcal N_f$, then there are a finite set $I \subseteq X$, $U_x \in \mathcal N_{f(x)}$ for $x \in I$ such that $\prod_{x\in I} U_x \times Y^{X\setminus I} \subseteq V$. By (2) for each $x \in I$ there is an $G_x \in D$ with $G_xx \subseteq U_x$, choose $G \in D$ with $G \subseteq \bigcap_{x \in I} G_x$, then $Gx \subseteq U_x$ for each $x \in I$, that is $G\subseteq \prod_{x \in X} U_x \times Y^{X\setminus I} \subseteq V$, as wished.

share|improve this answer
    
Yes, I understand. Thank you very much, and thank Sleziak again. –  Popopo Jul 8 '12 at 15:18
    
I did not understand :( please explain more so it will sort out my problem... –  user49974 Nov 19 '12 at 14:13
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.