Post History
#2: Post edited
by
ronno
·
2024-07-31T02:24:19Z (5 months ago)
This is not possible. Suppose there is such a $(P, S)$. Considering continuous maps to $\{0, 1\}$, there must be a clopen set $C \subset S$ such that there is no clopen $C' \subset P$ with $C = C' \cap S$.Now let $\kappa < \lambda$ be infinite cardinalities bigger than $|P|$ and consider the co-$\kappa$ topology on a set $X$ of size $\lambda$. Then $X$ is connected but subsets of size $\le \kappa$ are discrete. Now consider $x_1e x_2 \in X$ and the map $f : S \to X$ given by $f(C) = x_1$, $f(S \setminus C) = x_2$. This has no extension $g : P \to X$ since $x_1$ would be clopen in $f(P)$ and $C' = f^{-1}(x_1)$ would satisfy $C' \cap S = C$.In short, $(P, S)$-connectedness can only detect "connectedness of size $\le |P|$".
- This is not possible. Suppose there is such a $(P, S)$. Considering continuous maps to $\{0, 1\}$, there must be a clopen set $C \subset S$ such that there is no clopen $D \subset P$ with $C = D \cap S$.
- Now let $\kappa < \lambda$ be infinite cardinalities bigger than $|P|$ and consider the co-$\kappa$ topology on a set $X$ of size $\lambda$. Then $X$ is connected but subsets of size $\le \kappa$ are discrete. Now consider $x_1
- e x_2 \in X$ and the continuous map $f : S \to X$ given by $f(C) = x_1$, $f(S \setminus C) = x_2$. This has no continuous extension $g : P \to X$ since $x_1$ would be clopen in $g(P)$ and $D = g^{-1}(x_1)$ would satisfy $D \cap S = C$.
- In short, $(P, S)$-connectedness can at best detect "connectedness at size $\le |P|$".
#1: Initial revision
by
ronno
·
2023-05-04T13:28:00Z (over 1 year ago)
This is not possible. Suppose there is such a $(P, S)$. Considering continuous maps to $\{0, 1\}$, there must be a clopen set $C \subset S$ such that there is no clopen $C' \subset P$ with $C = C' \cap S$.
Now let $\kappa < \lambda$ be infinite cardinalities bigger than $|P|$ and consider the co-$\kappa$ topology on a set $X$ of size $\lambda$. Then $X$ is connected but subsets of size $\le \kappa$ are discrete. Now consider $x_1 \ne x_2 \in X$ and the map $f : S \to X$ given by $f(C) = x_1$, $f(S \setminus C) = x_2$. This has no extension $g : P \to X$ since $x_1$ would be clopen in $f(P)$ and $C' = f^{-1}(x_1)$ would satisfy $C' \cap S = C$.
In short, $(P, S)$-connectedness can only detect "connectedness of size $\le |P|$".