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Is there a limit ordinal $\kappa_0$ with $\kappa_0 \lt 2^{\aleph_0}$ and such that for every limit ordinal $\lambda$ with $\kappa_0\leq \lambda\lt 2^{\aleph_0}$ there is a connected $T_2$-space $X_\lambda$ with the following property?

$\lambda$ is the smallest ordinal such that $X_\lambda$ contains no subset isomorphic to $\lambda$ with the order topology

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    $\begingroup$ Do you know if every countable ordinal embeds into every non-trivial connected $T_2$ space? $\endgroup$ – James Hanson Sep 20 '18 at 17:14
  • $\begingroup$ Good question @JamesHanson. Even if we take everyone's darling space $\mathbb{R}$ with the Euclidean topology, I think it is not hard to see that $\omega \cdot \omega$ can be embedded into $\mathbb{R}$, but I am not sure about larger limit ordinals. So I do not know the smallest ordinal that cannot be embedded into $\mathbb{R}$ even. But this is certainly known, I'll try and Google it $\endgroup$ – Dominic van der Zypen Sep 20 '18 at 19:06
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    $\begingroup$ I'm fairly sure every countable ordinal can be embedded in $\mathbb{R}$. $\endgroup$ – James Hanson Sep 20 '18 at 19:34
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    $\begingroup$ @JamesHanson: That's correct. (You can prove it via transfinite induction. I don't have a reference, but I encourage you to give it a try -- it's a good exercise.) $\endgroup$ – Will Brian Sep 20 '18 at 20:00
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The answer is no: if $\lambda$ is larger than $\omega^2$ and if $X$ contains $\lambda+\omega$ then it also contains $\lambda+\omega+\omega$. To see this observe that $\lambda+1$ is homeomorphic with $\lambda+(\omega+1)+(\omega+1)$: simply take the first two copies of $\omega+1$ and move then to the end. So by contraposition: if $X$ does not contain $\lambda+\omega+\omega$ then it also does not contain $\lambda+\omega$, so that $\lambda+\omega+\omega$ is never `the smallest'. This does suggest, however, a modification of the question: look at indecomposable $\lambda$s.

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