From 60abe2db240e1b03746fe29ff219948fa8063960 Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 16:38:43 -0400 Subject: [PATCH 1/7] Upgrade ~locally connected to ~has open components --- spaces/S000107/properties/{P000041.md => P000234.md} | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) rename spaces/S000107/properties/{P000041.md => P000234.md} (85%) 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. From 02e32e504527b317956b0f4137e4aa2da6343513 Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:09:48 -0400 Subject: [PATCH 2/7] box product is slsc --- spaces/S000107/properties/P000229.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) create mode 100644 spaces/S000107/properties/P000229.md diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md new file mode 100644 index 0000000000..ac688dd6d0 --- /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 +$\mathbb{R}^\infty = \{y : y^n = 0\text{ for all but finitely many }n\}$. 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. From fb6bd125deaf880c403a1e56b97f4616cbd69a08 Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:14:04 -0400 Subject: [PATCH 3/7] Fix brackets --- spaces/S000107/properties/P000229.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md index ac688dd6d0..d4a3b29d75 100644 --- a/spaces/S000107/properties/P000229.md +++ b/spaces/S000107/properties/P000229.md @@ -9,15 +9,15 @@ refs: 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 +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]. +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 +Then because the restrictions of $F$ to each $\mathbb{R}^n \times [0, 1]$ are continuous and valued in $\mathbb{R}^n$, it follows that $F$ is continuous. From b04f97a8742d329f0e3ff62f3110c381b21ce82f Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:16:59 -0400 Subject: [PATCH 4/7] Adjustments 2 --- spaces/S000107/properties/P000229.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md index d4a3b29d75..ee4f5534b1 100644 --- a/spaces/S000107/properties/P000229.md +++ b/spaces/S000107/properties/P000229.md @@ -11,13 +11,14 @@ refs: 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}. +$\mathbb{R}^\infty := \{y : y^n = 0\text{ for all but finitely many }n\}$. +Because {P199|P229}, we may argue that $\mathbb{R}^\infty$ is +contractible. 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 $n$. +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 and valued in $\mathbb{R}^n$, it follows that $F$ is continuous. From b7346092cdd5d82bbc46cad6c0182fbfe92802ca Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:23:39 -0400 Subject: [PATCH 5/7] Fix formatting --- spaces/S000107/properties/P000229.md | 5 ++--- 1 file changed, 2 insertions(+), 3 deletions(-) diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md index ee4f5534b1..663682a980 100644 --- a/spaces/S000107/properties/P000229.md +++ b/spaces/S000107/properties/P000229.md @@ -12,10 +12,9 @@ refs: Since {S107|P86}, it is enough to show that the path-component of $0$ is {P229}. By {{mathse:5012784}}, this component equals $\mathbb{R}^\infty := \{y : y^n = 0\text{ for all but finitely many }n\}$. -Because {P199|P229}, we may argue that $\mathbb{R}^\infty$ is -contractible. +Then it is enough to argue that $\mathbb{R}^\infty$ is contractible ([Explore]([π-Base, Search for `contractible + ~Semilocally simply connected`](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. -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}}, From 95120c12e88f7a8d6bc364685ebec1f7b7f887f0 Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:26:35 -0400 Subject: [PATCH 6/7] Formatting 2 --- spaces/S000107/properties/P000229.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md index 663682a980..9988ed7096 100644 --- a/spaces/S000107/properties/P000229.md +++ b/spaces/S000107/properties/P000229.md @@ -12,7 +12,7 @@ refs: 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]([π-Base, Search for `contractible + ~Semilocally simply connected`](https://topology.pi-base.org/spaces?q=contractible+%2B+%7ESemilocally+simply+connected))). +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, From 3434e0065caf385e9c99a9981aa1327f385be193 Mon Sep 17 00:00:00 2001 From: Geoffrey Sangston Date: Fri, 10 Apr 2026 22:29:15 -0400 Subject: [PATCH 7/7] Fix --- spaces/S000107/properties/P000229.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/spaces/S000107/properties/P000229.md b/spaces/S000107/properties/P000229.md index 9988ed7096..ca8b0bc5ea 100644 --- a/spaces/S000107/properties/P000229.md +++ b/spaces/S000107/properties/P000229.md @@ -19,5 +19,5 @@ By {{mathse:3961052}}, the subspace topology on $\mathbb{R}^\infty$ coincides wi 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 and valued in $\mathbb{R}^n$, it follows +Then because the restrictions of $F$ to each $\mathbb{R}^n \times [0, 1]$ are continuous, it follows that $F$ is continuous.