Another proof that convergence in product topology $\prod\limits_{[0,1]}\mathbb{R}$ is pointwise convergence In a certain textbook I have a proof showing that convergence in product topology $\prod\limits_{[0,1]}\mathbb{R}$ is equivalent to pointwise convergence. Firstly, $\prod\limits_{[0,1]}\mathbb{R}$ is just the set of arbitrary functions $f:[0, 1] \rightarrow \mathbb{R}$. Then the implication that a sequence of functions $f^{(n)}$ converges to $f$ if $f^{(n)}(x)$ converges to $f(x)$ starts with defining a map $g: T \rightarrow \prod\limits_{[0,1]}\mathbb{R}, 1/n \mapsto f^{(n)}, 0\mapsto f$, with the domain being defined as the subspace $T$ consisting of a sequence $1/n$ and its limit value $0$; that is, $T=\{1/n \lvert n\in \mathbb{N}\} \cup {0}$. The following argument is what confuses me:

If the sequence of functions converges pointwise, then the composition $\pi_x g$ is continuous for all $x$ since then the image of the sequence $1/n$ converging to $0$ (and all other sequences in $T$ converging to $0$) converges to $f(x)$.

It is not clear for me why exactly continuity is implied. Firstly, nothing is stated about the topology of $\mathbb{R}$. I cannot assume that it is metric, and I cannot even assume that it is a first-countable space. It is strangely assumed that $0$ is the limit of $1/n$, and this shouldn't be true for some arbitrary topology on $\mathbb{R}$. Even if I assumed that $\mathbb{R}$ here is a first-countable space, the argument above would only show that the function $\pi_x g$ is continuous at $0$ (using the definition of sequential continuity).
I could finish the proof by assuming that $\mathbb{R}$ is first-countable and recalling that projections are open maps. So first-countability implies local continuity of $\pi_x g$ at point $0$. Then we can look at an arbitrary open neighbourhood $O$ of $f$ in product space. Since $\pi_x (O) = O'$ is an open neighbourhood of $f(x)$ in $\mathbb{R}$. Preimage of $O'$ under $\pi_x g$ is open and it is equal to the preimage of $O$ under $g$. This shows that $g$ is continuous at $0$ too, and since continuity preserves limits, the image of $(1/n) \subset T$ under $g$, that is $(f^{(n)})$, converges in product space.
But the proof in the textbook claims the general continuity of $\pi_x g$ which implies continuity of $g$ by the universal property of induced topology. In particular, the universal property doesn't impose that inducing maps are open.
I thus have two questions:

*

*Is there an implicit assumption about the topology of $\mathbb{R}$? If not, then why exactly the argument is true?

*Does the argument try to show continuity only at $0$ or does it somehow really says that $\pi_x g$ is continuous at every point? Is the argument really about sequential continuity or is there some other definition that implies continuity here that I'm missing?

 A: If nothing is stated to the contrary and a topology is used on $\Bbb R$ it must be the standard topology (from the metric or open intervals etc). This is common practice. But we don't really even need to know that:
If a sequence of functions $f^{(n)}$ converges to some function $f \in X:=\prod_{[0,1]} \Bbb R$ then it certainly converges pointwise: fix $x \in [0,1]$ and let $O$ be any open neighbourhood of $f(x)$. Then as the product topology is defined so that all projections: $\pi_x: X \to \Bbb R, x \in [0,1]$ are continuous we know that $\pi_x^{-1}[O]$ is an open neighbourhood of $f$ in $X$. As we assumed that $f^{(n)} \to f$ in $X$ we have some $N$ so that for all $n \ge N$ we have $f^{(n)}(x) \in \pi_x^{-1}[O]$ so that $f^{(n)}(x) \in O$ for all $n \ge N$ and it follows that $f^{(n)}(x) \to f(x)$ and this holds for any $x$ so $f^{(n)}$ converges pointwise to $f$.
The converse is also true: suppose $f^{(n)}$ converges pointwise for all $x \in [0,1]$ and let $O$ be a basic open neighbourhood of $f$ in $X$. By the definition of the product topology $$O=\bigcap_{i=1}^m \pi_{x_i}^{-1}[O_i]$$
where each $O_i$ is an open neighbourhood of $f(x_i)$ in $\Bbb R$ (whatever the topology) and we have finitely many such coordinates $\{x_1, \ldots, x_m\}\subseteq [0,1]$.
Then for each $i=1,\ldots,m$ we can find $N_i$ so that for all $n \ge N_i$ we have $f^{(n)}(x_i) \in O_i$, by pointwise convergence at $x_i$. Now define $N = \max_{i=1}^m N_i$ and if $n \ge N$ we have $n \ge N_i$ for all $i$ and so $$f^{(n)} \in \bigcap_{i=1}^m \pi_{x_i}^{-1}[O_i] = O$$ And as $O$ was an arbitrary basic open neighbourhood of $f$ we have $f^{(n)} \to f$ in $X$.
You see that the argument just uses continuity of the projections (for one direction) and the special form of basic product open sets for the other direction. Nothing really specific about the topology on $\Bbb R$ is assumed; it holds for whatever topology we pick on it.

The $T = \{\frac1n\mid n\} \cup \{0\}$ in your text is chosen as a special "test space" ( suspect that's why $T$ was chosen as its symbol, it's an idea one sees more often). It has the property that when $T$ has its usual topology (as I suppose it to be the case, and your text assumes it too):

Lemma: A function $h: T \to X$, where $X$ is any space is continuous iff $h(\frac1n) \to h(0)$ in $X$.

One direction is easy to see: if $h$ is continuous, $\frac1n \to 0$ in $T$ implies the condition $h(\frac1n) \to h(0)$ immediately (sequential continuity). And if the condition holds and $O$ is open in $X$, then $O'=h^{-1}[O]$ is open: any point of the form $\frac1m$ in $O'$ is automatically an interior point as $\{\frac1m\}$ is open in $T$. If $0 \in O'$ then $h(0) \in O$ and so there exists an $N$ (by the condition) so that for $n \ge N$, $h(\frac{1}{n}) \in O$, and it follows that $\{0\} \cup \{\frac1n\mid n \ge N\} \subseteq O'$ and the former set is an open neighbourhood of $0$ in $T$. So $0$ is an interior point too and $h^{-1}[O]$ is open and $h$ is continuous as $O$ was arbritary (not again, no info needed on the image space topology at all).
So in the proof that the text uses they seem to implicitly assume that you "see" this lemma right away.
And if then $f^{(n)} \to f$ pointwise for all $x \in [0,1]$, then fix $x \in [0,1]$ and it's clear that $$(\pi_x \circ g)(\frac1n) = \pi_x(f^{(n)}) = f^{(n)}(x) \to f(x) = \pi_x(g(0)) = (\pi_x \circ g)(0)$$ and the fact about $T$ then applies that $\pi_x \circ g: T \to \Bbb R$ is continuous. As this holds for all $x \in [0,1]$ the universal property for maps into the product $X$, implies that $g$ is continuous and this implies (by sequential continuity of $g$ applied to $\frac1n$ again) that $f^{(n)} \to f$ in $X$.
So your textbook proof works swimmingly provided that you see the easy lemma that I made explicit above. It's a tricky way to apply the universal property to see something that can also easily be seen using properties of its standard base, as we saw at the start.  I suppose it's good to have both tools at your disposal, and apply whatever is convenient for a given problem.
