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There was discussion of differences in box topology and product topology here.

From one answer, one question came to my mind.

Let $X$ be a non-discrete topological space.

Consider the diagonal map $f:X\rightarrow X\times X \times \cdots$ (say countable times product of $X$ with itself), $f(x)=(x,x,x,\ldots)$.

If we put product topology on $X\times X\times \cdots$ then $f$ is continuous (am I right?)

Question: Is it always true that for box topology on codomain of $f$, the map will never be continuous? If not, to make it non-continuous, what topology on $X$ should satisfy?

(In the link shared above, the example with $X=\mathbb{R}$ and usual topology on it illustrates that with box topology on $\mathbb{R}^{\mathbb{N}}$, the diagonal map is not continuous. )

Maths Rahul
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  • You are correct that the diagonal map $X \to X^{\mathbb N}$ is continuous when the latter space has the product topology, this is because, by the very definition of product topology, a map to a product is continuous iff every component is continuous. The same map is continuous when the codomain is endowed with the box topology if and only if every $G_\delta$ set of $X$ is open. I don't know if this is useful at all. – arnett Mar 01 '22 at 18:34

1 Answers1

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In the product topology the diagonal map $f: X \to X^{\Bbb N}$ is continuous, because $\pi_n \circ f = 1_X$ for all $n$ and the identity is continuous (universal continuity property of products).

If $x$ has a countable local base and $X$ is $T_1$, then $f$ will be continuous in the box topology on $\Bbb R^{\Bbb N}$ iff $x$ is an isolated point: we then have $U_n$ open in $X$ so that $\{x\} = \bigcap_n U_n$ and $f^{-1}[\prod_n U_n] = \bigcap_n U_n = \{x\}$ and so that singleton must already be open. In general you want for continuity of $f$ in that box situation that all $G_\delta$'s are open (as in the cocountable topology e.g.). So it can be continuous but often is not.

Henno Brandsma
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