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2 hours ago history edited Clement Yung CC BY-SA 4.0
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14 hours ago comment added Will Brian At one point in time, $\mathfrak d < \mathfrak a$ would have been a good answer to this question. Shelah's original proof of this inequality had $\mathfrak d >\!> \aleph_\omega$, with no obvious way of bringing the size down. But then later Shelah found a non-obvious way of doing it (template forcing) and found a model with $\aleph_2 = \mathfrak d < \mathfrak a = \aleph_3$. It is still an open problem whether $\aleph_1 = \mathfrak d < \mathfrak a$ is possible.
22 hours ago history asked Clement Yung CC BY-SA 4.0