pointwise convergence of a filter on $\mathbb{R}^\mathbb{R}$ In my topology lecture we have defined pointwise convergence for filters on function spaces, say $\mathbb{R}^\mathbb{R}$. A filter $\varphi$ on $\mathbb{R}^\mathbb{R}$ converges pointwise to $g:\mathbb{R}\to\mathbb{R}$ if for $x\in \mathbb{R}$ the filter generated by $\{F(x): F\in\varphi\}$ (where $F(x):=\{f(x): f\in F\}$) converges to $g(x)$.
I have no intuition what this means. Can someone breath life into this definition?
 A: Consider a sequence $\sigma=\langle f_n:n\in\Bbb N\rangle$ of elements of $\Bbb R^{\Bbb R}$; we say that this sequence converges pointwise to $g:\Bbb R\to\Bbb R$ if for each $x\in\Bbb R$, the coordinate sequence $\langle f_n(x):n\in\Bbb N\rangle$ converges to $g(x)$ in $\Bbb R$. Let’s look at that in a more general light.
As a set, $\Bbb R^{\Bbb R}$ is the Cartesian product of $|\Bbb R|$ copies of $\Bbb R$. When we look at $f(x)$ for some $f\in\Bbb R^{\Bbb R}$, we’re looking at the projection of $f$ on the $x$ factor of this product. If we let $R_x=\Bbb R$ for each $x\in\Bbb R$, then $\Bbb R^{\Bbb R}$ is simply $\prod_{x\in\Bbb R}R_x$, and for each $y\in\Bbb R$ we can define the projection map
$$\pi_y:\Bbb R^{\Bbb R}=\prod_{x\in\Bbb R}R_x \to R_y(=\Bbb R):f\mapsto f(y)\;.$$
From this point of view the coordinate sequence $\langle f_n(x):n\in\Bbb N\rangle$ is just the projection of $\sigma$ on the $x$-factor: it’s $\sigma_x=\langle\pi_x(f_n):n\in\Bbb N\rangle$. Thus, we say that $\sigma$ converges pointwise to $g$ if and only if for each $x\in\Bbb R$, the projection $\sigma_x$ of $\sigma$ on the $x$ factor converges to $g(x)$.
Pointwise convergence for filters on function spaces is a generalization of this idea. Instead of the sequence $\sigma$, we now have some filter $\varphi$ on $\Bbb R^{\Bbb R}$. Each $F\in\varphi$ is a subset of $\Bbb R^{\Bbb R}$, i.e., some set of functions from $\Bbb R$ to $\Bbb R$. Equivalently, it’s a subset of the Cartesian $\prod_{x\in\Bbb R}R_x$. As such it has a projection on each factor of this product: for each $x\in\Bbb R$ the projection of $F$ on $R_x$ is
$$\pi_x[F]=\{f(x):f\in F\}\;.$$
Let $\mathscr{B}_x=\{\pi_x[F]:F\in\varphi\}$; this is a collection of subsets of $R_x=\Bbb R$. When we projected the sequence $\sigma$ to $R_x$, we automatically got a sequence in $R_x$, i.e., a sequence of real numbers. Unfortunately, the projection $\mathscr{B}_x$ of the filter $\varphi$ on $R_x$ is not necessarily a filter. It is, however, a filter base: 


*

*for any $B_0,B_1\in\mathscr{B}_x$ there is a $B\in\mathscr{B}_x$ such that $B\subseteq B_0\cap B_1$. (This is an easy consequence of the fact that $\varphi$ is a filter, and I’ll leave its proof to you.)


This means that $\mathscr{B}_x$ generates a unique filter $\varphi_x$ on $R_x$ in the following way: let
$$\mathscr{B}_x'=\left\{\bigcap\mathscr{A}:\mathscr{A}\subseteq\mathscr{B}_x\text{ is finite}\right\}\;,$$
the closure of $\mathscr{B}_x$ under finite intersections, and set $\varphi_x=\{A\subseteq R_x:\exists B\in\mathscr{B}_x'(B\subseteq A)\}$, the closure of $\mathscr{B}_x'$ under supersets. You can easily check that $\varphi_x$ is a filter on $R_x=\Bbb R$.
Now we can say that the filter $\varphi$ on $\Bbb R^{\Bbb R}$ converges pointwise to $g\in\Bbb R^{\Bbb R}$ if and only if for each $x\in\Bbb R$, the filter $\varphi_x$ generated by the projection of $\varphi$ on the $x$ factor converges to $g(x)$. It really is exactly the same basic idea as pointwise convergence of sequences of functions.
