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visits member for 3 years, 10 months
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14h
comment In set theory, is there a name for a function which maps the empty set to zero and all the others to one?
$0^{0^S}$ -- wonderfully concise and intransparent.
19h
revised Continuous image relation on topological spaces
non-Hausdorff is the main point, not "finite".
1d
comment Freiling's Axiom of Symmetry Concretized
To prove "my" version from "your" version, just apply "your" version to the function $g(x):= f(x) \setminus \{x\}$.
1d
comment Freiling's Axiom of Symmetry Concretized
If you demand that $x\notin f(x)$ for all $x$, then the free set $F$ will have the property $y\notin f(z)$ for all $y,z\in F$. If you do not demand this, you get $y\notin f(z)$ only for all distinct $y,z\in F$.
2d
answered Freiling's Axiom of Symmetry Concretized
2d
answered Continuous image relation on topological spaces
2d
comment Continuous image relation on topological spaces
I am not sure if the empty set is considered a topological space, but in the context of this question it certainly should not be.
Mar
24
reviewed Close Probability or odds of something happening
Mar
24
awarded  Nice Answer
Mar
24
answered Connectedness in the language of path-connectedness
Mar
24
comment Connectedness in the language of path-connectedness
Very long lines exist. Let $\kappa$ be any cardinal (viewed as an ordinal). For each $\beta$ in $\kappa$ add a copy of the unit interval between $\beta$ and $\beta+1$, plus a point $\infty$. The resulting linear order is dense and complete. But if $C$ has cardinality smaller than $\kappa$, then any continuous image of $C$ that contains $0$ and $\infty$ will not be onto, hence not connected.
Mar
24
comment The product of the power and the natural number in the short interval
I do not understand the role of $a$ and $n$ here. Can't you just write $A$ for $a^n$?
Mar
24
comment Mal'cev “rational equivalence” and model theory
Concerning the model-theoretical point of view: The two structures use different languages, so formally the formulas satisfied in one structure are not the same as those satisfied in the other. But the assumption of "rational equivalence" implies that there are natural translations in both directions.
Mar
24
comment Mal'cev “rational equivalence” and model theory
Contrariwise. "rational equivalence" of two algebras implies that their clones are isomorphic. It is the other direction that is not necessarily true.
Mar
23
answered Mal'cev “rational equivalence” and model theory
Mar
23
comment Is $\mathcal{P}(\omega)/fin$ with the interval topology a connected space?
Nice! In Steen-Seebach's "Counterexamples in topology" such spaces where any two nonempty open sets intersect are called hyperconnected.
Mar
23
comment Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$?
Related question: mathoverflow.net/questions/89306/…
Mar
23
reviewed Approve hilbert-spaces tag wiki excerpt
Mar
17
answered On surjections, idempotence and axiom of choice
Mar
17
comment Surjective marriages
Yes iff $M$ is finite.