Let $\kappa$ be an infinite cardinal. Does there exist a topology $\tau\_{\kappa+1}$ on $\kappa+1$ such that for any topological space $(X,\tau)$ with $|X|=\kappa$ the following statement is true?



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> > $(X,\tau)$ is connected if and only if for all $x\_0, x\_1\in X$ there is a continous, injective map $f:\kappa+1 \to X$ such that $f(0) = x\_0$ and $f(\kappa) = x\_1$.
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(This is a follow-up to [Connectedness in the language of path-connectedness](https://mathoverflow.net/questions/200876/connectedness-in-the-language-of-path-connectedness).)



"Universal" connected spaces
No, there is not, even if you do not require $f$ to be injective. To see this, let $F$ be any ultrafilter on $\kappa$, and consider the following space $X\_F$. The underlying set of $X\_F$ is $\kappa\cup \{x\_0,x\_1\}$, with topology generated by the sets $\{\alpha\}$ for $\alpha\in\kappa$, $\{x\_0\}\cup\kappa$, and $\{x\_1\}\cup A$ for any $A\in F$. Note that $X\_F$ is connected, as it is easy to see that any clopen set containing $x\_0$ must be all of $X\_F$.


Now suppose the topology $\tau\_{\kappa+1}$ you ask for exists; let $f:\kappa+1\to X\_F$ be such that $f(0)=x\_0$ and $f(\kappa)=x\_1$. Let $F'$ be any ultrafilter other than $F$; I claim that $f$ is not continuous as a map $\kappa+1\to X\_{F'}$. Indeed, if it were, let $A\in F$ be such that $\kappa\setminus A\in F'$. Then $$f^{-1}(\{x\_1\})=f^{-1}(\{x\_1\}\cup A)\cap f^{-1}(\{x\_1\}\cup\kappa\setminus A)$$ would be open in $\kappa+1$. But $\{x\_1\}$ is closed in $X\_F$, so $f^{-1}(\{x\_1\})$ is closed and hence clopen in $\kappa+1$. But there clearly can be no clopen set that separates $0$ and $\kappa$ in $\kappa+1$, so this is a contradiction.


Thus each function $f:\kappa+1\to\kappa\cup\{x\_0,x\_1\}$ can be continuous as a map to $X\_F$ for at most one ultrafilter $F$. There are $2^{2^\kappa}$ ultrafilters on $\kappa$ and only $2^\kappa$ such functions $f$, so this is impossible.


