Cardinality of the set of all paths in the infinite complete infinitary tree - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:13:17Z http://mathoverflow.net/feeds/question/25623 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25623/cardinality-of-the-set-of-all-paths-in-the-infinite-complete-infinitary-tree Cardinality of the set of all paths in the infinite complete infinitary tree Halfdan Faber 2010-05-23T04:36:28Z 2012-01-31T17:54:44Z <p>The cardinality of the set of all root paths in the infinite complete binary tree is equal to the cardinality of the Continuum. The same holds true for k-ary trees for any finite k. But what is the case for k infinite? </p> http://mathoverflow.net/questions/25623/cardinality-of-the-set-of-all-paths-in-the-infinite-complete-infinitary-tree/25628#25628 Answer by Kevin Ventullo for Cardinality of the set of all paths in the infinite complete infinitary tree Kevin Ventullo 2010-05-23T06:06:01Z 2010-05-23T07:38:28Z <p>Assuming your path has countable length, the set of all paths in a \$k\$-ary tree will have cardinality \$k^{\aleph_0}\$. Indeed, at each step you have \$k\$ choices, and there are \$\aleph_0\$ steps (think of a path as a function from \$\mathbb{N}\$ to \$[k]\$).</p> http://mathoverflow.net/questions/25623/cardinality-of-the-set-of-all-paths-in-the-infinite-complete-infinitary-tree/25639#25639 Answer by Haim for Cardinality of the set of all paths in the infinite complete infinitary tree Haim 2010-05-23T08:47:45Z 2010-05-23T08:47:45Z <p>You may be also interested in the following paper by Shelah: <a href="http://shelah.logic.at/files/589.pdf" rel="nofollow">http://shelah.logic.at/files/589.pdf</a></p> <p>In this paper (section 2) he defines the more general notion of the "tree revised power" of two cardinals k1, k2 as the supremum on the number of k2-branches of trees with k1 nodes. He then proves that certain inequalities involving the tree revised power have some interesting consequences in pcf theory.</p>