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Let's call a topological space $X$ connected iff it can not be represented as a disjoint union of two non-empty open sets.

Now, suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Those two variants of connectivity are connected in the following ways:

If $X$ is $Y$-connected and $Y$ is connected then $X$ is also connected.

Proof: Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Converse is however generally false - for example, $\{(t, \sin(\frac{1}{t})|t \in \mathbb{R})\} \cup \{(0, 0)\}$ is connected, but not $[0;1]$-connected.

If $Y$ is not connected, then all topological spaces are $Y$-connected.

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected, Q.E.D.

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Suppose $x_0, x_1 \in X$. Thes $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now suppose $\alpha$ is a cardinal. let's call a topological space $Y$ of cardinality $\alpha$ $\alpha$-universally connected iff a topological space of cardinality $\leq \alpha$ being $Y$-connected is equivalent to it being connected (equivalently, $Y$ is connected and all connected spaces of cardinality at most $\alpha$ are $Y$-connected)

My question is:

For what $\alpha$ do $\alpha$-universally connected spaces exist?

This question is a follow up of that one: Do universally connected spaces exist?

Chain Markov
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  • I'm not sure but it seems to me that, if there is an $\alpha$-universally connected space, then there is a connected space $Y$ of cardinality $\alpha$ such that every nonempty connected space of cardinality $\le\alpha$ is a continuous image of $Y$. The idea is to sew together a whole bunch ($\alpha^2$) of copies of your $\alpha$-connected space. – bof Jan 08 '21 at 13:10
  • Trivial starting point: this holds for $\alpha = 1$. – user3482749 Jan 08 '21 at 13:23

1 Answers1

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Partial answer.

For a given cardinal $\alpha$ the collection of all connected topological spaces with cardinality at most $\alpha$ (up to homeomorphism) is a set. Let $C$ be that set. Then $Y=\prod C$ is $\alpha$-universally connected. There's even stronger property here: every connected topological space of cardinality up to $\alpha$ is a quotient of $Y$.

Except that $Y$ itself may be (and probably is) of bigger than $\alpha$ cardinality, I'm not sure.

freakish
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