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?



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> $\lambda$ is the smallest ordinal such that $X\_\lambda$ contains no subset isomorphic to $\lambda$ with the [order topology](https://en.wikipedia.org/wiki/Order_topology)
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Embedding ordinals with the order topology into connected $T_2$-spaces
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.


