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 Apr 23 comment “Downward closed” relation on a poset Downward closed sets are sometimes called "order ideals". Apr 17 awarded Enlightened Apr 17 comment Cardinality of connected Hausdorff topologies My answer suggests that a more interesting question might have been "how many locally connected Hausdorff spaces...". But I am pretty sure that the answer would stay the same. Apr 17 comment Cardinality of connected Hausdorff topologies Call the original space $A$, the "strings" (copies of the open unit interval or connected space) $B$, and let $c$ be the point where are strings are glued together. As Jeremy Rickard points out, the very many topologies on $A$ induce very many topologies on $A\cup B\cup \{c\}$, and the number of homeomorphism classes stays "very large" (i.e., $2^{2^\kappa}$). -- However, at least in the case $\kappa\ge 2^{\aleph_0}$, it is clear that each autohomeomorphism must map $B$ onto $B$ (all neighborhoods are $\simeq (0,1)$), and $c$ to $c$, hence $A$ onto $A$. Apr 17 awarded Nice Answer Apr 17 revised Cardinality of connected Hausdorff topologies deleted 12 characters in body Apr 17 answered Cardinality of connected Hausdorff topologies Apr 8 comment $\mathfrak{ufo}$: An unidentified combinatorial cardinal characteristic of the continuum? Welcome to math overflow! Nice to see you here! Mar 22 comment Reference request, proof about well-ordering of ordinals "A system where you can quantify over predicates" describes the language of second order logic. This description does not tell you which proofs you allow. One way (the cheap way, but still good enough for most of mathematics) is to translate second order statements into first order statements in the language of ZFC, and then use your favorite first order proof system (plus ZFC axioms, or more) to analyse these statements. Mar 21 revised A weak kind of fixed point tag: lattice-theory Mar 21 comment Is there a particular field that cannot be proven to have an algebraic closure in ZF? (A generalisation of @JulianRosen's comment:) For any ordered (i.e., linearly ordered) field $K$, the ordered finite extensions form a directed system. (The embeddings between them are canonical, because of the order.) Their naturally defined limit (without AC, I think) is a real closed field. Adjoin an imaginary unit $i$ to get an algebraically closed field. Mar 20 comment Reference request, proof about well-ordering of ordinals There is no canonical notion of "second order proof". Which notion do you have in mind? Mar 2 answered Why should we believe in the axiom of regularity? Feb 22 revised Bidi: A new cardinal characteristic of the continuum? polished Polish Feb 17 comment When can you canonically extend an ultrafilter after forcing? Do you have an example of a forcing that has this property? Feb 15 comment Has anything ever been done with the set $\{1,2,3,4,\ldots\}$ equipped with the operation $a \oplus b = a+b-1$ and the usual notion of multiplication? The title of your question is misleading, because the structure $(\{1,2,\ldots\,\},\oplus)$ is isomorphic to the well-known structure $(\{0,1,2,\ldots\, \},+)$. You are interested in $\{1,2,\ldots\},\oplus,\cdot)$. Jan 15 answered Is there an uncountable Borel almost disjoint family? Dec 29 comment Unconventional types of induction Thanks for fixing it. Dec 17 revised Are undecidable consequences of Con recursively enumerable? \\, -> \, (TeX) Dec 11 comment Under what conditions can we put a complete norm on a linear subspace of a separable Banach space? If you drop the separability assumption, a similar fact is true: An infinite-dimensional vector space over the reals of dimension $\kappa$ admits a complete norm if and only if $\kappa = \kappa^{\aleph_0}$. (Arthur Kruse, Badly incomplete normed linear spaces. Mathematische Zeitschrift, 1964.)