# Embedding ordinals with the order topology into connected $T_2$-spaces

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

• Do you know if every countable ordinal embeds into every non-trivial connected $T_2$ space? – James Hanson Sep 20 '18 at 17:14
• 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 – Dominic van der Zypen Sep 20 '18 at 19:06
• I'm fairly sure every countable ordinal can be embedded in $\mathbb{R}$. – James Hanson Sep 20 '18 at 19:34
• @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.) – Will Brian Sep 20 '18 at 20:00

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.