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It is easy to show that $\mathbb{N}$ with the cofinite topology is not path connected and that any set with cardinality $\geq 2^{\aleph_0}$ equipped with the cofinite topology is in fact path connected.

what about cardinalities $\aleph_0<\alpha<2^{\aleph_0}$ (under the assumption that such exist obviously)?

If $\alpha $ is path connected then any cardinality $\geq \alpha$ is also, so an interesting direction would be trying and checking whether $\aleph_1$ is path connected (I have no clue about how one can even start checking this).

Noam Zimhoni
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  • It's classical that every nonempty Hausdorff compact perfect space has cardinal $\ge c$. The image of a non-constant path in a Hausdorff compact space satisfies these assumptions, and hence has cardinal $\ge c$. Hence, every Hausdorff compact space of cardinal $<c$ is totally path-disconnected. – YCor Jul 05 '19 at 19:21
  • (No need of compactness in the last sentence of my last comment: the consequence is: every Hausdorff space of cardinal $<c$ is totally path-disconnected.") – YCor Jul 05 '19 at 19:50
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    @YCor: The space in question isn't Hausdorff, though. In fact, I think that the cofinite topology on some cardinal $\alpha$ should be path connected provided that $\alpha$ is $\geq$ the cardinal $\acute{\mathfrak{n}}$ from this question: https://mathoverflow.net/questions/285780/are-the-sierpi%c5%84ski-cardinal-acute-mathfrak-n-and-its-measure-modification?noredirect=1&lq=1. The idea is that if ${X_\xi ,:, \xi < \alpha }$ is a partition of $[0,1]$ into closed sets, then the mapping that sends $X_\xi$ to $\xi$ is continuous (when $\alpha$ has the cofinite topology). – Will Brian Jul 05 '19 at 20:08
  • A path connected space is also connected by injective paths, so if it has more than one point, it has at least c points (but is it true for T1spaces ?) – Pietro Majer Jul 05 '19 at 20:22
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    @PietroMajer I only know that your first statement is true for Hausdorff spaces. The line with origin doubled is T1 and path-connected, but not arc-connected. – Wojowu Jul 05 '19 at 20:40
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    @WillBrian In fact, this is if and only if, isn't it? – Emil Jeřábek Jul 06 '19 at 05:45

1 Answers1

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A continuous non-constant function from $[0,1]$ into $X$ with the cofinite topology exists iff $[0,1]$ has a partition into $\le |X|$ many disjoint closed non-empty subsets.

This question discusses the options for the cardinality of such a partition. One of the conclusions is that under $\textrm{MA}(\omega_1)$ we have that a set of size $\aleph_1$ in the cofinite topology is (connected and ) not path-connected, while of course under CH such a set is path-connected. So the case $\aleph_1$ is undecidable under ZFC.

Henno Brandsma
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