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Jul 6, 2019 at 14:01 vote accept Noam Zimhoni
Jul 6, 2019 at 6:09 answer added Henno Brandsma timeline score: 5
Jul 6, 2019 at 5:45 comment added Emil Jeřábek @WillBrian In fact, this is if and only if, isn't it?
Jul 5, 2019 at 20:40 comment added Wojowu @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.
Jul 5, 2019 at 20:22 comment added Pietro Majer 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 ?)
Jul 5, 2019 at 20:08 comment added Will Brian @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).
Jul 5, 2019 at 19:50 comment added YCor (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.")
Jul 5, 2019 at 19:21 comment added YCor 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.
Jul 5, 2019 at 18:15 review First posts
Jul 5, 2019 at 18:15
Jul 5, 2019 at 18:13 history asked Noam Zimhoni CC BY-SA 4.0