Does this "jumping-ahead" ordinal function exist? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:00:39Z http://mathoverflow.net/feeds/question/99060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99060/does-this-jumping-ahead-ordinal-function-exist Does this "jumping-ahead" ordinal function exist? Manny Reyes 2012-06-07T20:25:54Z 2012-06-07T22:56:59Z <p>While working on a project in operator algebras with a <a href="http://mathoverflow.net/users/10368/chris-heunen" rel="nofollow">collaborator (and fellow MO user)</a>, we are able to successfully complete a transfinite induction assuming that the following has an affirmative answer.</p> <blockquote> <p><strong>Question:</strong> Let $\delta$ be a cardinal, considered as an <a href="http://en.wikipedia.org/wiki/Ordinal_number#Initial_ordinal_of_a_cardinal" rel="nofollow">initial ordinal</a>, so that it is equal to the set of all ordinals of cardinality strictly less than $\delta$. Does there exist a nondecreasing function $\phi \colon \delta \to \delta$ such that, for all ordinals $1 \leq \lambda &lt; \delta$, there exists $\gamma &lt; \lambda$ such that $\phi(\gamma) \geq \lambda$?</p> </blockquote> <p>If it happens to matter, we are only concerned with the case where $\delta$ is a limit cardinal. Any reasonably (transfinitely) constructive approach to writing down such $\phi$ seems to quickly run into issues of <a href="http://en.wikipedia.org/wiki/Ordinal_notation" rel="nofollow">ordinal notation</a> that are beyond our expertise. If an abstract existence argument is available, we will certainly still be happy. On the other hand, we fear that this may somehow depend on large cardinal issues. </p> http://mathoverflow.net/questions/99060/does-this-jumping-ahead-ordinal-function-exist/99062#99062 Answer by Nik Weaver for Does this "jumping-ahead" ordinal function exist? Nik Weaver 2012-06-07T21:03:07Z 2012-06-07T22:40:58Z <p>Not unless $\delta$ has countable cofinality (e.g., $\delta = \aleph_\omega$). This will fail for $\delta = \aleph_1$, for example. Let $\phi: \delta \to \delta$ be any increasing function and recursively define $\lambda_0 = 0$ and $\lambda_{n+1} = \phi(\lambda_n)+1$. Since $\phi$ is increasing, the sequence $(\lambda_n)$ is increasing, and since $\delta$ has uncountable cofinality we have $\lambda = \sup \lambda_n &lt; \delta$. However, for any $\gamma &lt; \lambda$ we must have $\gamma &lt; \lambda_n$ for some $n$, so that $\phi(\gamma) \leq \lambda_{n+1} &lt; \lambda$.</p> <p>(If $\delta$ has countable cofinality it's easy. For instance, if $\delta = \aleph_\omega$ then we define $\phi$ by letting $\phi(\lambda) = \aleph_{n+1} + \lambda$ for $\aleph_n \leq \lambda &lt; \aleph_{n+1}$.)</p>