Introducing pointwise convergence of functions as convergence with respect to some metric

Question

I tried to explain the notion of pointwise/uniform convergence of functions to undergrad students whose only background is metric spaces.

Let $\{f_n:I\rightarrow \mathbb{R}\}$ be a sequence of continuous/bounded functions. We say $(f_n)$ converge uniformly/pointwise to a function $f:I\rightarrow \mathbb{R}$ if some conditions are satisfied.

I thought it would be easier for students to understand the difference if I show the difference between uniform and pointwise convergence as convergence in two different metrics.

Let $X$ be a set. Fix a metric $d:X\times X\rightarrow \mathbb{R}$ in $X$. Let $a:\mathbb{N}\rightarrow X$ be a sequence in $X$, which we denote, for convinience, by $(a_n)$. We say that the sequence $(a_n)$ in $X$ converges to an element $a\in X$, if, given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $d(a_n,a)<\epsilon$ for all $n\geq N$.

In this situation $f_n\in \mathbb{R}^{I}$ for each $n\in \mathbb{N}$. There is a metric on the subset, of bounded functions on $I$, of the set $\mathbb{R}^I$. The metric is given by $d_{\infty}(f,g)=\sup_{x\in I} |f(x)-g(x)|$ for $f,g$ bounded in $I$.

So, with this metric, $(f_n)$ converge to $f$ given $\epsilon>0$, there exists $n\in \mathbb{N}$ such that, $d(F_n,f)<\epsilon$. By definition, $d_{\infty}(f_n,f)=\sup_{x\in I} |f_n(x)-f(x)|$. So, given $\epsilon>0$, there exists $N\in \mathbb{N}$ such that $|F_n(x)-f(x)|<\epsilon$ for all $x\in I$.

This is precisely the definition of uniform convergence of functions. A sequence of functions $(f_n)$ in $B(I,\mathbb{R})$ is said to be uniformly convergent to an element $f$ in $B(I,\mathbb{R})$ if it is convergent to $f$ in the metric $d_{\infty}$.

Question : Can I introduce pointwise convergence in similar manner (with out talking about topology)? I want to say $(f_n)$ converges to $f$ pointwise in $B(I,\mathbb{R})$ if and only if $(f_n)$ converges to $f$ in $B(I,\mathbb{R})$ with metric $d$ on $B(I, \mathbb{R})$. Does such a metric exits?

To talk about uniform convergence, I could use the sup metric on $B(I,\mathbb{R})$. To talk about pointwise convergence, what metric should I use? I have no hope to get such a metric. Just want to confirm as I have learnt this long back.

I guess the easiest way to see this is that the topology of pointwise convergence isn't first countable? — WoolierThanThou, Oct 23 '19 at 11:48
You might present this to the students: Consider the set of all functions from $\Bbb N$ to $[0,1]$ with the metrics $d_1(f,g)=\sup {|f(n)-g(n)|: n\in \Bbb N}$ and $d_2(f,g)=\sum_{n\in \Bbb N}2^{-n}|f(n)-g(n)|.$ A sequence is $d_1$-convergent iff it is uniformly convergent. A sequence is $d_2$-convergent iff it is point-wise convergent... If $f_m$ is the characteristic function of ${j\in \Bbb N:j\le m}$ then $(f_m)_m$ converges point-wise, but not uniformly, to $1$. — DanielWainfleet, Oct 23 '19 at 14:17
sharing the reason for downvote might help to make this question better — Praphulla Koushik, Nov 06 '19 at 12:18

score 1 · Accepted Answer · edited Nov 25 '19 at 13:24

We may identify the set $B(I,\mathbb R)$ with a subset of the infinite product $P = \prod_{t\in I} \mathbb R_t$, where all $\mathbb R_t = \mathbb R$. In fact, each function $f : I \to \mathbb R$ corresponds to the point $(f(t))_{t\in I} \in P$.

Now give $P$ the product topology induced from its factors $\mathbb R_t$. Then a sequence $(f_n)$ converges pointwise to $f$ iff $(f_n)$ converges to $f$ with respect to this topology ("componentwise convergence").

Your question is therefore when $P$ is a metrizable space.

If $I$ is finite with $n$ elements, then $P$ can be identified with $\mathbb R^n$ and you get the desired metric. Also if $I$ is countable infinite $P$ turns out to be metrizable. See for example Show that the countable product of metric spaces is metrizable.

However, if $I$ is uncountable, then $P$ is not metrizable. In fact, it is easy to see that it is not first countable.

Thanks. I knew about topology.. It was about metric.. thanks for your answer.. — Praphulla Koushik, Oct 23 '19 at 12:19

HUO · Answer 2 · 2023-03-19T11:05:47.270

The accepted answer does not address the question. The question was about the existence of a metric that induces pointwise convergence, while the answer dealt with the fact that the "standard topology" inducing pointwise convergence is not metrizable. This simply does not rule out the existence of such a metric.

However, it is indeed true that no metric induces pointwise convergence of continuous functions (hence, in particular, measurable functions) on a closed interval. This was proved in the following paper.

Fort, M. K. jun., A note on pointwise convergence, Proc. Am. Math. Soc. 2, 34-35 (1951). ZBL0043.11303.

Introducing pointwise convergence of functions as convergence with respect to some metric

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