diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md
new file mode 100644
index 0000000000..ca8b0bc5ea
--- /dev/null
+++ b/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 
+$\mathbb{R}^\infty := \{y : y^n = 0\text{ for all but finitely many }n\}$. 
+Then it is enough to argue that $\mathbb{R}^\infty$ is contractible ([Explore](https://topology.pi-base.org/spaces?q=contractible+%2B+%7ESemilocally+simply+connected)).
+This 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 topology on $\mathbb{R}^\infty \times [0, 1]$ also coincides with 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.
diff --git a/spaces/S000107/properties/P000041.md b/spaces/S000107/properties/P000234.md
similarity index 85%
rename from spaces/S000107/properties/P000041.md
rename to spaces/S000107/properties/P000234.md
index ab470ec1d0..4627b12ef5 100644
--- a/spaces/S000107/properties/P000041.md
+++ b/spaces/S000107/properties/P000234.md
@@ -1,10 +1,10 @@
 ---
 space: S000107
-property: P000041
+property: P000234
 value: false
 refs:
 - mathse: 5012784
   name: Answer to "Is $\ell^\infty$ with box topology connected?"
 ---
 
-By {{mathse:5012784}} the connected component of an arbitrary point $x\in X$ is $A = \{y : y_n = x_n\text{ for all but finitely many }n\}$. Since $\text{int}(A) = \emptyset$, it follows that $x$ has no connected neighbourhoods.
+By {{mathse:5012784}} the connected component of an arbitrary point $x\in X$ is $A = \{y : y_n = x_n\text{ for all but finitely many }n\}$. Since $\text{int}(A) = \emptyset$, it follows that $A$ is not open.