Problem. Assume that a compact space $X$ can be written as the union $X=K\cup D$ of a compact metrizable subspace $K$ and a discrete subspace $D$. Does $D$ contain a non-trivial convergent sequence in $X$?

As shown by Ilya Bogdanov (Does every compact countable space contain a non-trivial convergent sequence?) the answer is affirmative if $K$ is at most countable.

  • $\begingroup$ What happens if $|D|=1$? $\endgroup$ – Ilya Bogdanov Nov 8 '16 at 10:16
  • $\begingroup$ Dear Ilya, it seems that I know the complete (positive) answer to my problem. I will try to write it down now. $\endgroup$ – Taras Banakh Nov 8 '16 at 10:45

Oh, sorry! I wrote this question and after some thinking found a (relatively simple) answer.

Consider the set $\mathcal P$ of pairs $(A,I)$ where $A$ is a non-empty closed subset of the compact metrizable space $K$ and $I$ is an infinite subset of $D$ such that every open neighborhood $U$ of $A$ in $X=K\cup D$ contains all but finitely many points of the set $I$. The set $\mathcal P$ is endowed with the partial order: $(A,I)\le (B,J)$ if $A\subset B$ and $I\setminus J$ is finite. Let $(A_\alpha,I_\alpha)_{\alpha<\kappa}$ be a maximal descreasing transfinite sequence of elements of $\mathcal P$ such that for any ordinals $\alpha<\beta<\kappa$ the set $A_\beta$ is a proper subset of $A_\alpha$. The second countability (in fact, the hereditary Lindelofness) of $K$ guarantees that the length $\kappa$ of such sequence is countable. Moeover, the ordinal $\kappa$ is successor. In the opposite case we can extend the transfinite sequence $(A_\alpha,I_\alpha)_{\alpha<\kappa}$ letting $A_\kappa=\bigcap_{\alpha<\kappa}A_\alpha$ and $I_\kappa$ be any pseudointersection of the countable tower $(I_\alpha)_{\alpha<\kappa}$. So, $\kappa=\lambda+1$ for some ordinal $\lambda$. We claim that $A_\lambda$ is a singleton, and hence $I_\lambda$ is a required non-trivial convergent sequence. If $A_\lambda$ is not a singleton, we can write it as the union $A_\lambda=B\cup C$ of two proper closed subsets of $A_\lambda$. By the maximality of the transfinite sequence $(A_\alpha,I_\alpha)_{\alpha\le\lambda}$, the set $B$ has an open neighborhood $U\subset X$ such that $J=I_\lambda\setminus U$ is infinite. Using the maximality once more, we can find an open neighborhood $V\subset X$ of $C$ such that $J\setminus V$ is infinite. Then $U\cup V$ is an open neighborhood of $A_\lambda$ in $X$ such that $I_\lambda\setminus (U\cup V)=J\setminus V$ is infinite, which contradicts the inclusion $(A_\lambda,I_\lambda)\in\mathcal P$.

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