1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $(A_\lambda,I_\lambda)\in\mathcal P$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.