Given a topological space $C$ and points $c\_0, c1\in C$ we say that a topological space $(X,\tau)$ is $(C,c\_0,c\_1)$-connected if and only if for all $x,y\in X$ there is $f:C\to X$ continuous with $f(c\_0) = x$ and $f(c\_1)=y$.


(In this language, path-connectedness equals $([0,1],0,1)$-connectedness.)


What is an example of a space $C$ and points $c\_0\neq c1\in C$ such that


1. $(C,c\_0,c\_1)$-connectedness implies path-connectedness, and
2. for every infinite cardinal $\kappa$ there is a topology on $\tau$ on $\kappa$ such that $(\kappa,\tau)$ is path-connected, but not $(C,c\_0,c\_1)$-connected


?



Stronger form of connectedness than path-connectedness
Let $C$ be the space nº74 ("double origin topology") in Steen & Seebach's *Counterexamples of Topology*, chosen because it is the only one listed there that is T2 and path-connected but not T3:



> 
> $C$ consists of the set of points of the plane $\mathbb{R}^2$ together with an additional point $0^\*$. Neighborhoods of points other than the origin $0$ and the point $0^\*$ are the usual open sets of $\mathbb{R}^2\setminus\{0\}$; as a basis fo neighborhoods of $0$ and $0^\*$, we take $V\_n(0) = \{(x,y) : x^2+y^2 < 1/n^2 \land y>0\} \cup \{0\}$ and $V\_n(0^\*) = \{(x,y) : x^2+y^2 < 1/n^2 \land y<0\} \cup \{0^\*\}$.
> 
> 
> 


In other words we have replaced the origin in the plane by an "upper origin" $0$ and a "lower origin" $0^\*$, the neighborhoods of the upper origin being sets containing an open half-disk centered at the origin, plus the upper origin itself, and similarly for the lower origin.


The space $C$ is Hausdorff, but not T2½ because $0$ and $0^\*$ do not have disjoint closed neighborhoods; in particular, it is not T3 or T3½. So there is no continuous function $C \to \mathbb{R}$ taking different values on $0$ and $0^\*$.


Let $c\_0 = 0$ and $c\_1 = 0^\*$. Then there is a path connecting $c\_0$ and $c\_1$: indeed, there is a path connecting $c\_0$ to, say, $(1,0)$, and one connecting $(1,0)$ to $c\_1$ (we can even find "arcs", i.e., injective paths, if we want). If we have a continuous function $C \to X$ taking $c\_0$ to $x$ and $c\_1$ to $y$, then right-composing it with the path just mentioned gives a path connecting $x$ and $y$: so $(C,c\_0,c\_1)$-connectedness implies path-connectedness. On the other hand, $\mathbb{R}$ is not $(C,c\_0,c\_1)$-connected because of what was said in the previous paragraph.


I think this answers the question, with the additional constraint that $C$ is Hausdorff and $c\_0 \neq c\_1$. (I just noticed that Simon Henry had the same idea in the comments.)


**However,** it doesn't answer the question that I think *should* have been asked, namely to also require $C$ itself to be $(C,c\_0,c\_1)$-connected (in the above example, it's pretty clear that it's not).


