arXiv · 0908.1942
I prove that a Hilbert space has the property that each of its dense (not necessarily closed) subspaces contains an orthoormal basis if and only if it is separable.
Ilijas Farah. 2009-08-15. Orthonormal bases of Hilbert spaces. https://arxiv.org/abs/0908.1942
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The Kolmogorov complexity of a string characterize how complex it is to describe the string. If every prefix of a real $x$ is more complex to describe than every prefix (of the same length) of real $y$, then it is seen as $x$ is more complex to describe than $y$. It is wondered if there is a real $x$ so that no other reals are strictly more complex (to describe) than $x$. The behavior of Kolmogorov complexity functions generated by reals (namely $n\mapsto$ the minimal description length of the real) is quite chaos. Therefore, it is widely believed that there are many reals that are maximally complex to describe. For instance, it is conjectured that all random enough reals have maximal $K$-degree. In this paper, it is shown that there is no real with maximal $K$-degree. Actually, for almost all real $x$, we can uniformly computably find another real whose $K$-degree is strictly above $x$.
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