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  1. I was wondering how "the topology of pointwise convergence" is defined on $Y^X$ where $X$ is a set and $Y$ is a topological space?

    Are there more than one topologies that can topologize pointwise convergence? If yes, which one is the natural/canonical one?

  2. I was wondering how "the uniformity of uniform convergence" is defined on $Y^X$ where $X$ is a set and $Y$ is a uniform space?

    Are there more than one uniform structures that can uniformize uniform convergence? If yes, which one is the natural/canonical one?

  3. In the first case, as many references point out, "the topology of pointwise convergence" on $Y^X$ is the product topology on $Y^X$. I happen to realize that "topology of pointwise convergence" is the minimum topology on $Y^X$ s.t. the evaluation at each $x \in X$ is a continuous mapping from $Y^X$ to $Y$.

    In the second case, as this source points out, "the uniformity of uniform convergence" on $Y^X$ is a uniform structure on $Y^X$ generated by the base defined as the collections of all pairs $(f,g) \in Y^X \times Y^X$ such that $(f(x), g(x)) \in V$ for all $x \in X$ and where $V$ runs through a base of entourages for $Y$. If I understand correctly, "the uniformity of uniform convergence" on $Y^X$ is the minimum uniform structure on $Y^X$ s.t. the evaluation at each $x \in X$ is a uniformly continuous mapping from $Y^X$ to $Y$.

    The above two equivalences are very similar in that they make all the evaluations of $Y^X$ at each $x \in X$ become morphisms wrt the structures on $Y$. I was wondering if the two equivalences are just two coincindences?

Thanks and regards!

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2 Answers 2

up vote 6 down vote accepted

As you note, the topology of pointwise convergence on $Y^X$ is simply the product topology on the product of $|X|$ copies of $Y$ indexed by $X$. For $x\in X$ let $e_x:Y^X\to Y:f\mapsto f(x)$ be the evaluation map at $x$, and let $\tau$ be the coarsest topology on $Y^X$ making each $e_x$ continuous. For each finite $F\subseteq X$ and open $V\subseteq Y$ let $$B(F,V)=\{f\in Y^X:f(x)\in V\text{ for each }x\in F\}\;;$$ the collection of all such $B(F,V)$ is a base $\mathscr{B}$ for $\tau$.

Suppose that $\tau\,'$ is a topology on $Y^X$ such that a net $\nu=\langle f_d:d\in D\rangle$ in $Y^X$ converges to $f\in Y^X$ in $\tau\,'$ iff $\langle f_d(x):d\in D\rangle\to f(x)$ for each $x\in X$. Since $\langle f_d(x):x\in D\rangle$ is just the image of $\nu$ under $e_x$, this says that each $e_x$ sends convergent nets with their limits to convergent nets with their limits, which implies that each $e_x$ is continuous with respect to $\tau\,'$ and hence that $\tau\subseteq\tau\,'$.

Suppose that $\tau\,'\supsetneqq\tau$, and let $U\in\tau\,'\setminus\tau$. Because $U\notin\tau$, there is an $f\in U$ such that for each $B(F,V)\in\mathscr{B}$, either $f\notin B(F,V)$, or $B(F,V)\nsubseteq U$. Let $\mathscr{F}$ be the family of non-empty finite subsets of $X$, let $\tau(Y)$ be the topology on $Y$, and let $\mathscr{D}=\mathscr{F}\times\tau(Y)$. For $\langle F,V\rangle,\langle G,W\rangle\in\mathscr{D}$ define $\langle F,V\rangle\le\langle G,W\rangle$ iff $F\subseteq G$ and $V\supseteq W$; then $\langle\mathscr{D},\le\rangle$ is a directed set.

Let $$\nu:\mathscr{D}\to Y^X:B(F,V)\mapsto g_{F,V}$$ be any net in $Y^X$ such that $g_{F,V}\in B(F,V)\setminus U$ whenever $f\in B(F,V)$. Fix $x\in X$, and let $V$ be any open nbhd of $f(x)$ in $Y$. If $\langle\{x\},V\rangle\le\langle F,W\rangle\in\mathscr{D}$, then $e_x(g_{F,W})=g_{F,W}(x)\in W\subseteq V$, so $\big\langle g_{F,V}(x):\langle F,V\rangle\in\mathscr{D}\big\rangle$ converges to $f(x)$ in $Y$. However, it’s clear that $\nu$ does not converge to $f$ in $\tau\,'$, since $\nu$ is not eventually in $U$. Thus, $\tau$ is the unique topology on $Y^X$ such that an arbitrary net $\langle f_d:d\in D\rangle$ in $Y^X$ converges to $f\in Y^X$ in $\tau\,'$ iff $\langle f_d(x):d\in D\rangle\to f(x)$ for each $x\in X$.


The coarsest uniformity on $Y^X$ making each evaluation map $e_x:Y^X\to Y:f\mapsto f(x)$ uniformly continuous is the uniformity of pointwise convergence. If $\mathscr{U}$ is the (diagonal) uniformity on $Y$, for each $x\in X$ and $U\in\mathscr{U}$ let

$$V_{x,U}=\left\{\langle f,g\rangle\in Y^X\times Y^X:\langle f(x),g(x)\rangle\in U\right\}\;;$$

then $\{V_{x,U}:x\in X\text{ and }U\in\mathscr{U}\}$ is a subbase for the uniformity of pointwise convergence on $Y^X$.

This is in general different from the uniformity of uniform convergence on $Y^X$: that has as a subbase the family $$\left\{\bigcap_{x\in X}V_{x,U}:U\in\mathscr{U}\right\}$$ and is finer than the uniformity of pointwise convergence.

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There are perfectly good books that explain all this: Kelley (Topology), Engelking (General Topology) among the standard ones.

But you answer question 1 in point 3: it is indeed the product topology on $Y ^ X$. It is unique if you want the following property : for all nets $(f_i)_{i \in I}$ (where $(I, \le)$ is a directed set, as usual) with values in $Y ^ X$, the net converges to $f \in Y^X$ iff for all $x \in X$, the net $(f_i(x))_{i \in I}$ converges to $f(x)$ (in $Y$, which is a topological space, so it makes sense to talk about convergence of nets there).

This defines on $Y^X$ a criterion to determine the convergence of nets (when we know the topology on $Y$). This means we know what the closed sets are (exactly those sets that are closed under net convergence), and thus the open sets (their complements). So knowing net convergence amount to the same thing as knowing the topology, if we verify that this net convergence definition (or criterion) is indeed topological, as in Kelly's 4 criteria, that you have asked about earlier. This is somewhat tricky. So what we do instead is define a topology directly on $Y^X$, the product topology, which is the smallest topology that makes all projections, which in this case are the evaluations $f \rightarrow f(x), x \in X$, continuous. We then prove that the property for nets actually holds, thereby showing it is indeed a topological convergence criterion (by definition, as we now know it comes from a topology). So the topology induced by the criterion is the product topology (it uniquely determines the topology if it is topological).

As to the uniformity question, the minimal uniformity that makes all projections/evaluations uniformly continuous is the uniformity of pointwise convergence, and the topology it induces is just the product topology. This construction is the direct analogue of the pointwise convergence one. Both are the category-theoretic product in their respective categories. The uniformity you describe is in general much finer.

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Nice that Brian actually did the proof I alluded to in full (that the product topology satisfies the pointwise convergence criterion). –  Henno Brandsma Feb 24 '13 at 8:02
    
Thanks! The two references are appreciated too. –  Tim Feb 24 '13 at 15:12

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