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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).

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Jul 5, 2019 at 19:21
  • $\begingroup$ (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.") $\endgroup$
    – YCor
    Commented Jul 5, 2019 at 19:50
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    $\begingroup$ @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: mathoverflow.net/questions/285780/…. 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). $\endgroup$
    – Will Brian
    Commented Jul 5, 2019 at 20:08
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    $\begingroup$ @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. $\endgroup$
    – Wojowu
    Commented Jul 5, 2019 at 20:40
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    $\begingroup$ @WillBrian In fact, this is if and only if, isn't it? $\endgroup$
    – Emil Jeřábek
    Commented Jul 6, 2019 at 5:45

1 Answer 1

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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.

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