If $X=[0,1]^{[0,1]}$ in product topology, then it is a compact Hausdorff space which is not second countable. However I think this space is infinite in (covering, inductive)dimension. Does there exist a compact Hausdorff space which is not second countable and finite in dimension?
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1$\begingroup$ For compact non-metrisable spaces ind, Ind and dim dimension functions can differ. Which do you mean? $\endgroup$– Henno BrandsmaCommented Oct 17, 2021 at 14:50
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$X=\omega_1 + 1$ is not second countable, compact and hereditarily normal, and has $\dim(X)=\text{Ind}(X)=0$. I think $0$ counts as finite...
$\{0,1\}^{[0,1]}$ also has the same dimension values and is compact and non-second countable.
answered Oct 17, 2021 at 14:52
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$\begingroup$ Thank you for the answer. Can you further explain what is $\omega_{1}+1$? I am not familiar with this space. $\endgroup$– Ken.WongCommented Oct 17, 2021 at 15:42
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1$\begingroup$ @Ken.Wong The space of all countable ordinals plus compactifying point. Munkres calls it $\overline{S_\Omega}$ IIRC $\endgroup$ Commented Oct 17, 2021 at 15:45