Consider the cartesian product $$X= ⨉_{r \in[0,1]} \ [0,\infty) = \{\text{all functions } f:[0,1] \rightarrow [0,\infty)\}$$
with the (product) topology generated by the projections
$$
S = \{p_r^{-1}(U) \colon r \in [0,1], U \subseteq [0,\infty) \text{ open}\}.
$$
Let $A \subseteq X$. How can I in general compute the closure of $A$ with respect to this topology?
Say for instance $A=\{f \in X \colon f >0\}$? Can anyone help me in the right direction?
I would like to get a better grasp on the product topologies!
Update:
I now think I have cracked how to do it for sets like $A$: We can use the fact that in the product topology
$$
\overline{⨉_{\alpha \in I} U_\alpha} = ⨉_{\alpha \in I} \overline{U_\alpha}.
$$
So we obtain that
$$
\overline{A} = \overline{\{f \in X \colon f >0\}} = \overline{⨉_{r \in(0,1]} \ (0,\infty)} = ⨉_{r \in(0,1]} \ \overline{(0,\infty)} = ⨉_{r \in(0,1]} \ [0,\infty) = X.
$$
I am wondering if this is the default method or there is something better? Say we for instance instead want to find the closure of as set like
$$
B=\{f \in X \colon \{ x \in [0,1] \colon f(x) >0 \} \text{ is finite } \}
$$
I can't really find any way to write this as a cartesian product since the condition of finite non-zero is a "local" property for each $f$. Is it possible or is there another way to tackle this? If we could write it as a cartesian product then we would be done by the above method.
Any thoughts or help is appreciated!
Best Answer
The method you've used for $A$ seems to be computing the sequential closure, so I don't think an analogous method for $B$ exists, as the sequential closure of $B$ is strictly smaller than the closure of $B$. In such a case, I only know how to show it from the basis of the product topology as follows:
Since $S$ is a subbasis for the product topology, we have the basis elements $$\bigcap_{r \in I}p_r^{-1}(U_r)$$ where $I\subset [0,1]$ is a finite index set and each $U_r \subseteq [0,\infty)$ is open. In particular, these basis elements are nonempty if and only if each $U_r$ is nonempty.
Suppose now we have a nonempty basis element in the given form. Then there is a function $f : [0,1] \to [0,\infty)$ so that $f(r) \in U_r$ and $f(x) = 0$ whenever $x \not\in I$. Therefore $f \in \bigcap_{r \in I}p_r^{-1}(U_r)$.
Additionally, we see that $\{x \in [0,1] : f(x) > 0\} \subseteq I$ is finite, so $f \in B$. Therefore $$B \cap \bigcap_{r \in I}p_r^{-1}(U_r) \neq \emptyset.$$
That is, $B$ is dense in $X$, so its closure is $X$.
As an additional problem that begs to be asked by the above: Show that the sequential closure of $B$ is sequentially closed but is not equal to $X$. In particular, $X$ is not a sequential space.