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| bio | website | mathoverflow.net/users/2000 |
|---|---|---|
| location | Hanover, NH | |
| age | 37 | |
| visits | member for | 2 years, 11 months |
| seen | 9 hours ago | |
| stats | profile views | 322 |
Mathematician and MathOverflow moderator.
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May 14 |
awarded | Enlightened |
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May 14 |
awarded | Nice Answer |
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May 12 |
accepted | A natural example in category theory |
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May 8 |
comment |
A natural example in category theory Oh! This is excellent! I think this has finite products but not all finite limits. Right? |
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May 7 |
comment |
A natural example in category theory This is not an explicit requirement but it is implied: if $X$ is itself inhabited then it is clearly isomorphic to a subobject of an inhabited object. |
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May 7 |
asked | A natural example in category theory |
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May 7 |
awarded | Caucus |
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Apr 26 |
comment |
What is Baire's zero-dimensional metric space? Additional context would help, but it's usually this one - en.wikipedia.org/wiki/Baire_space_(set_theory) |
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Apr 26 |
awarded | Informed |
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Apr 5 |
comment |
Uniform distribution with probability density function. Find the value of $k$. How many types of mathematicians are there? |
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Mar 27 |
awarded | Good Answer |
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Jan 7 |
comment |
What are the differences between rings, groups, and fields? That used to be the case but most authors today define a ring to have $1$. The unusual looking term rng is sometimes used for the concept without $1$. |
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Dec 27 |
comment |
Is empty set a proper subset of itself? $B \setminus A \neq \varnothing$ does not imply that $A$ is a subset of $B$. (But, if it is, then it is indeed a proper subset.) |
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Dec 5 |
answered | Questions about generalizations of the Principle of Dependent Choices |
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Dec 4 |
answered | Proofs whose length depends on the input |
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Sep 18 |
answered | Does every nonempty definable finite set have a definable member? |
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Aug 19 |
answered | Borel linear order cannot have uncountable increasing chain |
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Aug 19 |
comment |
Borel linear order cannot have uncountable increasing chain I had missed the link! The author cites Harrington and Shelah, who did prove the result I recalled earlier. I guess it would be best to check that reference. @William: No. There are no uncountable wellordered Borel chains at all so that weaker variant is vacuously true. |
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Aug 19 |
comment |
Borel linear order cannot have uncountable increasing chain This is true if chain is replaced by wellordered chain. In other words, a Borel linear order cannot contain a copy of $\omega_1$ or its reverse. I would guess that's what is meant, otherwise "increasing or decreasing" is not very meaningful. Where is this from? |
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Aug 16 |
comment |
What is actually “relatively consistent”? "If a system is not complete, then it is consistent." Seeing that complete usually means "proves $\sigma$ or $\lnot\sigma$ for every sentence $\sigma$" and that consistent usually means "does not prove every sentence $\sigma$," a system that is not complete must be consistent. There are variations but, in any case, I don't think your second sentence exactly says what you meant to say. |
