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What is the Topology of point-wise convergence? It has been stated in lectures but I am unfamiliar with it.

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3 Answers 3

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Let $F$ be a family of functions from a set $X$ to a space $Y$. $F$ might, for instance, be the set of all functions from $X$ to $Y$, or it might be the set of all continuous functions from $X$ to $Y$, if $X$ is a topological space. Each $f:X\to Y$ can be thought of as a point in the Cartesian product $Y^{|X|}$. To see this, for each $x\in X$ let $Y_x$ be a copy of the space $Y$. Then a function $f:X\to Y$ corresponds to the point in $\prod_{x\in X}Y_x$ whose $x$-th coordinate is $f(x)$, and of course $\prod_{x\in X}Y_x$ is just the product of $|X|$ copies of $Y$, i.e., $Y^{|X|}$.

The product $Y^{|X|}$ is a topological space with the product topology; $F\subseteq Y^{|X|}$, so $F$ inherits a topology from the product topology on $Y^{|X|}$. This inherited topology is the topology of pointwise convergence on $F$.

It can easily be shown that a sequence $\langle f_n:n\in\Bbb N\rangle$ in $F$ converges to some $f\in F$ in this topology if and only if for each $x\in X$, $\langle f_n(x):n\in\Bbb N\rangle$ converges to $f(x)$ in $Y$. (More generally, a net $\langle f_d:d\in D\rangle$ in $F$ converges to some $f\in F$ if and only if for each $x\in X$ the net $\langle f_d(x):x\in D\rangle$ converges to $f(x)$ in $Y$.) This is the reason for the pointwise in the name.

Very often $Y$ is $\Bbb R^n$ or $\Bbb C^n$ for some $n$, and $X$ is some topological space. The topological structure of $X$ has no bearing on the topology of pointwise convergence, though it may help to determine the set $F$ of functions under consideration (e.g., the continuous ones).

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    $\begingroup$ Brilliant! For the statement 'f_n is converges in the product topology iff f(x) converges in Y for each x in X' Initially I thought one would need uniform convergence in Y, but is it due to the product topology being defined to have open sets of the form, 'Cartesian product of only finitely many proper open sets'. Is this why this works? As then when considering convergence in the product topology when we take an open set, we will only require to look at finitely many of the x's corresponding to those proper open subsets. $\endgroup$
    – user58514
    Feb 2, 2013 at 23:15
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    $\begingroup$ @rustyracketman: Right. In the product topology you can restrict only finitely many coordinates at once, so while you can get convergence in each coordinate, you can’t get that convergence to occur at a uniform rate: that would require being able to restrict every coordinate, as in the uniform topology. $\endgroup$ Feb 2, 2013 at 23:18
  • $\begingroup$ And then, by convergence in Y we get the existence of an N(x) for each x (The N(x) such that all f_n(x) are contained in the corresponding open set of f(x), for any n>= N(x)) Then since we are only looking at finitely many x's we can take maximum of these N(x)'s. And thus f_n will be contained in the original open set that contained f, for all n>= Max{N(x)}. $\endgroup$
    – user58514
    Feb 2, 2013 at 23:20
  • $\begingroup$ @rustyracketman: Yes. $\endgroup$ Feb 2, 2013 at 23:21
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From Munkres' Topology, Second Edition, $\S$46:

Definition.   Given a point $x$ of the set $X$ and an open set $U$ of the space $Y$, let $$S(x,U)=\{f\mid f\in Y^X\hbox{ and }\ f(x)\in U\}.$$ The sets $S(x,U)$ are a subbasis for a topology on $Y^X$, which is called the topology of pointwise convergence (or the point-open topology).

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  • $\begingroup$ Can this be used to show the definition the other two have given? If so, will you give me a hint please? $\endgroup$
    – user58514
    Feb 2, 2013 at 19:26
  • $\begingroup$ Yup. Try deconstructing the definition in a metric space using open balls instead of arbitrary open sets. Then see if you can generalize your deconstruction to the general case. $\endgroup$ Feb 3, 2013 at 16:26
  • $\begingroup$ I find this the most intuitive definition, especially when considering finite intersections of these sabbatical open sets. $\endgroup$
    – ABIM
    Dec 14, 2022 at 5:17
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It's a topology on the functions of a space such that a sequence of functions $\phi_n$ converges to $\phi$ if and only if for all $x$ in the space the sequence of points $\phi_n(x)$ converges to $\phi(x)$.

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