# What is the Topology of point-wise convergence?

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

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).

• 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. – user58514 Feb 2 '13 at 23:15
• @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. – Brian M. Scott Feb 2 '13 at 23:18
• 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)}. – user58514 Feb 2 '13 at 23:20
• @rustyracketman: Yes. – Brian M. Scott Feb 2 '13 at 23:21

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 topolgoy of pointwise convergence (or the point-open topology).

• Can this be used to show the definition the other two have given? If so, will you give me a hint please? – user58514 Feb 2 '13 at 19:26
• 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. – Avi Steiner Feb 3 '13 at 16:26

The topology of point-wise convergence is a topology on a function space. Let $Z=Y^X$ be the space of functions from $X$ to $Y$. Then a sequence $(f_i)_{i\in \mathbb N}$ converges to a function $f$ if for all $x\in X$, the sequence $(f_i(x))$ converges to $f(x)$.

• Thankyou, why are they talking about THE topology of point-wise convergence then? – user58514 Feb 2 '13 at 19:24
• Because if you are given $X$ and $Y$, there is only one topology of pointwise convergence on $Y^X$. You can only talk about a topology on a space after you have what that space is. But beyond that, if only $X$ is given, then "functions on $X$" tends to refer to real valued functions on $X$, and so in lieu of everything being explicit, you can get away with specifying just $X$. Note that, unless you are restricting to continuous functions, the topology of point-wise convergence does not depend on the topology on $X$. – Aaron Feb 2 '13 at 19:44

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 $\phi(x)$.