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

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You mean $Y^X$ instead of $X^Y$, right? – martini Jul 6 '12 at 9:09
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
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|cite|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

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