# Does this generalization of path-connectedness also cover general connectedness?

I've got the following idea to generalise path-connectedness:

Given a topological space $P$ and a subspace $S$, a space $X$ is $(P,S)$-connected if every continuous function $f:S\to X$ can be extended to a continuous function $g:P\to X$.

The usual path-connectedness is obtained using $P=[0,1]$ with the usual topology and $S=\{0,1\}$. Since every function $f:\{0,1\}\to X$ is continuous, $f$ just picks two arbitrary points, and then $g$ is a path between those points.

Using the closed unit disc as $P$ and the unit circle as $S$ one also obtains whether all path-connected components are simply connected.

Now I wonder if there is also a choice of $P$ and $S$ that recovers general connectedness, that is the nonexistence of clopen sets other than the empty set and the full space.

## 1 answer

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 $\le |P|$".

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