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Feb 6, 2019 at 10:33 comment added Santi Spadaro Ok, I've just added it.
Feb 6, 2019 at 6:02 comment added Dap @SantiSpadaro: oh, wow. That should be an answer so it's more visible
Feb 4, 2019 at 15:11 comment added Santi Spadaro $2^\kappa$ for uncountable $\kappa$ is a ZFC example of an almost sequential non-sequential space. To see why it's almost sequential, let $x$ be any point in $2^\kappa$ and let $D=\{y \in 2^\kappa: |\{\alpha < \kappa: x(\alpha) \neq y(\alpha) \}| < \aleph_0 \}$. Then $D$ is a sequential (even Fréchet-Urysohn) dense subspace of $2^\kappa$ which contains $x$.
Feb 3, 2019 at 5:45 history edited Dap CC BY-SA 4.0
typo
Feb 3, 2019 at 5:21 comment added Dap @TarasBanakh: I think I've weakened the assumption to $\mathfrak{d}=\aleph_1.$
Feb 3, 2019 at 5:17 history edited Dap CC BY-SA 4.0
added argument with weaker assumption d=aleph_1
Feb 2, 2019 at 14:11 history edited Dap CC BY-SA 4.0
the example construction of f_\alpha made no sense
Feb 2, 2019 at 9:28 vote accept Taras Banakh
Feb 2, 2019 at 9:28 comment added Taras Banakh @Dab Thank you for the corrections. Now it indeed works! Now it would be nice to weaken CH to something like $\mathfrak u\le\mathfrak d$.
Feb 2, 2019 at 8:32 comment added Dap That should be "neighborhood subbase" in the last comment. Also I did spot one $\subseteq^*$ reversed in the "Specifically,.." sentence, fixed now.
Feb 2, 2019 at 8:10 history edited Dap CC BY-SA 4.0
the subset was the wrong way round when referring to the construction in Jech
Feb 2, 2019 at 7:43 history edited Dap CC BY-SA 4.0
fix to previous edit
Feb 2, 2019 at 7:12 comment added Dap @TarasBanakh: Thanks for the comments! The sets $U_{n,\alpha}$ are not increasing in $\alpha$ - they just form a subbase of $*.$ The sequence $S_\alpha$ is $\supseteq^*$-increasing i.e. decreasing modulo finite sets. I've fixed the ordinal notation.
Feb 2, 2019 at 7:01 history edited Dap CC BY-SA 4.0
fix successor ordinal notation + a slightly too strong assumption in the last point
Feb 1, 2019 at 10:22 comment added Taras Banakh And the main problem is with your definition of the topology at $*$: as $\alpha$ icreases so do the sets $U_{n,\alpha}$ but it should be vice-versa. If you replace $\le$ by $\ge$ in the definition of $U_{n,\alpha}$, then you will obtain the standard compact metrizable topology on $(\omega\times\{\omega\})\cup\{*\}$ and this will not allow you to prove that your space is not sequential.
Feb 1, 2019 at 10:17 comment added Taras Banakh Thank you for another answer. I have some remarks: (1) it seems that the sequences $(S_\alpha)$ and $(f_\alpha)$ can be constructed under the assumtion $\mathfrak p=\mathfrak c$ (which is weaker than CH and follows from MA, actually it is equivalently to some version of MA); (2) usually by $\kappa^+$ the successor cardinal to $\kappa$ is denoted. So, your $\omega^+$ is a bit misleading; your can write $\omega+1$ or $\bar\omega$ instead.
Feb 1, 2019 at 7:16 history edited Dap CC BY-SA 4.0
added 35 characters in body
Feb 1, 2019 at 7:11 comment added Dap @TarasBanakh: Sorry, I hadn't thought that through. I've now come up with a different construction using CH.
Feb 1, 2019 at 7:05 history undeleted Dap
Feb 1, 2019 at 7:05 history edited Dap CC BY-SA 4.0
new argument
Jan 29, 2019 at 12:57 history deleted Dap via Vote
Jan 29, 2019 at 12:23 comment added Taras Banakh Thank you for the answer, but your third condition contradicts the second one. In fact, the second is true but the third is not.
Jan 29, 2019 at 12:05 history undeleted Dap
Jan 29, 2019 at 12:05 history edited Dap CC BY-SA 4.0
added 74 characters in body
Jan 29, 2019 at 11:58 history edited Dap CC BY-SA 4.0
added 153 characters in body
Jan 29, 2019 at 11:44 history deleted Dap via Vote
Jan 29, 2019 at 11:18 history answered Dap CC BY-SA 4.0