box product is slsc

Author: GeoffreySangston

spaces/S000107/properties/P000229.md

@@ -0,0 +1,23 @@
+---
+space: S000107
+property: P000229
+value: true
+refs:
+- mathse: 3961052
+  name: Answer to "Is the weak topology on $\mathbb{R}^{\infty}$ the same as the box topology?"
+- mathse: 833227
+  name: Answer to "(Certain) colimit and product in category of topological spaces"
+---
+
+Since [S107|P86], it is enough to show that the path-component of $0$
+is [P229]. By {{mathse:5012784}}, this component equals 
+$\mathbb{R}^\infty = \{y : y^n = 0\text{ for all but finitely many }n\}$. We argue that $\mathbb{R}^\infty$ is
+contractible, since [P199|P229].
+
+The claim follows once we argue that $F : \mathbb{R}^\infty \times [0, 1] \to \mathbb{R}^\infty$, $(x, t) \mapsto tx$, is continuous.
+By {{mathse:3961052}}, the subspace topology on $\mathbb{R}^\infty$ coincides with the weak topology,
+where a set $U \subset \mathbb{R}^\infty$ is open if and only if $U \cap \mathbb{R}^n$ is open for
+each $\mathbb{R}^n := \{x \in \mathbb{R}^\infty : x^m = 0\text{ if } m > n\}$. By {{mathse:833227}},
+the product of $\mathbb{R}^\infty \times [0, 1]$ also has the weak topology, where again a set $U \subset \mathbb{R}^\infty \times [0, 1]$ is open if and only if $U \cap (\mathbb{R}^n \times [0, 1])$ is open for each $n$.
+Then because the restrictions of $F$ to each $\mathbb{R}^n \times [0, 1]$ are continuous, it follows
+that $F$ is continuous.