Is it consistent that $\frak{d} < 2^{\aleph_0}$?

Question

Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$. We write $f <^* g$ if there is $N\in\omega$ such that $f(n) < g(n)$ for all $n>N$. A set $D\subseteq \omega^\omega$ is said to be dominating if for all $f\in \omega^\omega$ there is $g\in D$ such that $f <^* g$. Set $$\frak{d} = \textrm{min}\{|\mathrm{D}|: \mathrm{D}\subseteq \omega^\omega \textrm{ and } \mathrm{D} \textrm{ is dominating}\}.$$

http://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum — Simon Thomas, Dec 22 '14 at 14:04
See the reference to Hechler in this answer: http://mathoverflow.net/a/29626 — François G. Dorais, Dec 22 '14 at 14:43

score 12 · Answer 1 · answered Dec 22 '14 at 14:08

12

Use the fact that adding any number of random reals does not increase the dominating number since every new real is dominated by a ground model real.

answered Dec 22 '14 at 14:08

Guest1245

121
2

Avshalom · Accepted Answer · 2014-12-22T14:23:03.620

4

Theorem 5.1 in Eric van Douwen's paper "The integers and topology" (Handbook of Set-theoretic Topology) is an old reference for a positive answer to your question and will provide a fuller explanation of the proof mentioned by Guest1245. There are newer references too: see the relevant chapter in L. Halbeisen, Combinatorial Set Theory. With a gentle introduction to forcing.

edited Dec 22 '14 at 14:23

answered Dec 22 '14 at 14:12

Avshalom

2,121

Is it consistent that $\frak{d} < 2^{\aleph_0}$?

2 Answers2