User arthur fischer - MathOverflow

Two versions of "absolutely ccc"

2012-03-23T17:04:33Z

I have recently been slogging my way through Shelah's "Large continuum, oracles". Essentially from the start there has been a question needling me which I cannot seem to answer.

In the paper, Shelah says that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after forcing with any ccc notion.
Elsewhere, I have seen it defined that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after any forcing. (This would be indestructibly ccc from Bartoszyński-Judah.)

Any forcing having the Knaster property is absolutely ccc (in the strong sense), and MA$_{\aleph_1}$ implies that all ccc forcings have the Knaster property. Thus, it is consistent that the two are equivalent.

Do these two versions of absolute ccc-ness provably coincide?

Answer by Arthur Fischer for How to see such space is Lindelof?

2013-02-20T08:31:43Z

Note that the open subsets of (what I will denote by) $\mathbb{R}_B$ are of the form $U \cup A$ where $U \subseteq \mathbb{R}$ is open in the usual topology, and $A \subseteq B$ is arbitrary.

Suppose that $\{ U_i \cup A_i : i \in I \}$ is an open cover of $\mathbb{R}_B$. Note that there is a countable $I_0 \subseteq I$ such that $\bigcup_{i \in I_0} U_i = \bigcup_{i \in I} U_i$. Next note that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq B$ is closed (in $\mathbb{R}$) and is therefore countable, so there is a countable $I_1 \subseteq I$ such that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq \bigcup_{i \in I_1} A_i$.

Answer by Arthur Fischer for Existence of weakly compact cardinals

2012-10-09T09:19:44Z

As weakly compact cardinals are in particular (strongly) inaccessible, it follows from Gödel's Second Incompleteness Theorem that ZFC cannot prove the implication "Con(ZFC) implies Con (ZFC + $\exists$weakly-compact )." (Unless, of course, if ZFC is itself inconsistent, at which point this is all a lot of bunk).

(The linked Wikipedia article outlines the basic reasoning.)

Any argument for the consistency of "ZFC + $\exists$weakly-inaccessible" must therefore transcend ZFC. (But, as mentioned by Asaf, no contradictions have been found thus far been under the assumption of the existence of weakly-compact cardinals.)

Answer by Arthur Fischer for A function that is defined everywhere but has unknown values

2012-09-14T13:03:48Z

Kind of a cheat, but define $f : \mathbb{N} \to \mathbb{N}$ so that $f(n)$ is the initial position (to the right of the decimal point) of the first occurance of $n$ consecutive $5$s in the the decimal expansion of $\pi$, and $0$ if such does not exist.

So $f(1) = 4$, $f(2) = 130$, $f(3) = 177$, $f(4) = 24,466$, etc. I don't think I'm saying too much by claiming that as it is unknown whether $\pi$ is normal, we do not know if $f(n)$ is non-zero for all $n$.

EDIT: I honestly didn't see the almost identical answer in the comments above before posting this. It has thus been made CW.

Answer by Arthur Fischer for countable topological spaces of uncountable weight

2012-08-21T05:17:18Z

Another classical example is the Arens-Fort Space.

Let $X = \omega \times \omega$. For each $A \subseteq X$ and $n \in \omega$ we let $A_n = \{ m \in \omega : (n,m) \in A \}$ denote the $n$th section of $A$.

Topologise $X$ by taking each point of $X \setminus \{ (0,0) \}$ to be isolated, and let $U \subseteq X$ be a neighbourhood of $(0,0)$ iff it contains $(0,0)$ and all but finitely many sections of $U$ are co-finite.

Clearly this space is T$_1$. If $F, E \subseteq X$ are disjoint closed sets, then one, say $E$, does not contain $(0,0)$. So $E$ is clopen, and $X \setminus E$ is an open set including $F$ which is disjoint from $E$. Thus $X$ is normal.

To show that $X$ is not second countable, it suffices to show that there is no countable base at $(0,0)$. If $\{ U^{(i)} \}_{i \in \omega}$ is any family of open neighbourhoods of $(0,0)$, we inductively define a sequence of pairs of natural numbers $\{ (n_i,m_i) \}_{i \in \omega}$ so that:

$n_i > n_{i-1}$ is such that $U^{(i)}_{n_i}$ is non-empty (say $n_{-1} = 0$); and
$m_i \in U^{(i)}_{n_i}$.

Then $V = X \setminus \{ ( n_i , m_i ) : i \in \omega \}$ is an open neighbourhood of $(0,0)$, and by construction $U^{(i)} \not\subseteq V$ for all $i$.

Comment by Arthur Fischer

2013-03-14T10:43:31Z

What sort of axiomatization are you thinking of for NBG/MK? I think the most common axiomatizations include (or imply) Limitation of Size, which says that a class is proper iff it admits an injection from V, and from which you then get Global Choice.

Comment by Arthur Fischer

2013-02-20T09:57:50Z

@John: Do you mean my characterisation of the open subsets of $\mathbb{R}_B$? (Which follows from the fact that the topology generated by the usual open subsets of $\mathbb{R}$ and the singletons from $B$.) Or that there is a countable $I_0$? (Which follows from the fact that $\mathbb{R}$ is second-countable, and thus hereditary Lindelöf.)

Comment by Arthur Fischer

2012-08-20T17:24:11Z

Didn't Hilbert himself produce a constructive proof a few years after his non-constructive one?

Comment by Arthur Fischer

2012-03-24T08:01:49Z

@saf: See e.g., Jech (3rd ed.), Theorem 16.21, p.277. The proof actually gives the slightly stronger result that MA$_{\aleph_1} $implies that all ccc posets have precalibre $\aleph_1$.

Comment by Arthur Fischer

2012-03-24T06:36:15Z

To further Joel's point (or confusion) I believe that the usual construction of a Hausdorff gap yields an indestructible gap. It thus seems that this answer would contradict what I've stated in the question about the two formulations of absolute ccc-ness being consistently equivalent.