Stronger form of connectedness than path-connectedness

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

?

• $(\{*\},*,*)$ ?. – Simon Henry Mar 31 '16 at 12:01
• More seriously, unless I'm mistaken, If $C$ is not functionally separated (more precisely if $c_0$ and $c_1$ cannot be separated by continuous functions) then $[0,1]$ is not going to be $C$-connected and conversely if $C$ is functionally separated then every path connected space will be $C$-connected. So you are looking for examples of path-connected non functionally separated spaces. Lots of non Hausdorff example, and I'm sure books like "counterexamples in topology" contains lots of Hausdorff examples. – Simon Henry Mar 31 '16 at 12:12
• ... and obviously it will also implies that you don't have many $C$-connected spaces: a functionally separated space will always be "$C$ totally disconected". – Simon Henry Mar 31 '16 at 12:21
• Right, the one-space example... I changed the question a bit to evade this example – Dominic van der Zypen Mar 31 '16 at 14:36

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).

• With regard to your last remarks: In fact it would already be nice to have an example where $C$-connectedness is a non trivial notion with (i.e. there is as least one non trivial space that is $C$-connected) – Simon Henry Mar 31 '16 at 14:41
• @Gro-Tsen very good point, can you ask the question with your additional requirement? – Dominic van der Zypen Mar 31 '16 at 14:59
• @SimonHenry: There are still the indiscrete spaces, but you're right, it's even worse than I thought. – Gro-Tsen Mar 31 '16 at 15:00
• @DominicvanderZypen: I'll rather let you do that, but I suggest you give the matter further thought to decide what the best question should be, because it's not entirely clear. You probably want to require $C$ to be $(C,c_0,c_1)$-connected and $c_0$ and $c_1$ (hence any two points) to be connected by a path and sprinkle some magic "Hausdorff" dust in various places, but I'm not sure what else. – Gro-Tsen Mar 31 '16 at 15:06
• Thanks @Gro-Tsen - will think about it and post a new question in a week or so – Dominic van der Zypen Apr 1 '16 at 6:47