| bio | website | math.cas.cz/~jerabek |
|---|---|---|
| location | Prague | |
| age | 35 | |
| visits | member for | 1 year, 4 months |
| seen | 17 hours ago | |
| stats | profile views | 15 |
I am a researcher at the Institute of Mathematics of the Czech Academy of Sciences. I work in the field of mathematical logic, specifically proof complexity (mainly subsystems of bounded arithmetic, but also propositional proof complexity) and nonclassical logics (admissible rules of modal, superintuitionistic, and other propositional logics).
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Mar 6 |
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Topological characterization of the closed interval $[0, 1]$. A cutpoint is a point that when removed doesn’t leave the remainder still connected, right? |
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Feb 5 |
awarded | Yearling |
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Feb 5 |
answered | Two forms of Beth's theorem? |
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Jan 18 |
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Metrizable group @jspecter: The quotient of any topological group by a closed normal subgroup is Hausdorff. |
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Jan 14 |
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The roots in a finite field The criterion for solvability can be written more consisely as $a^{(q-1)/\gcd(m,q-1)}=1$. |
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Jan 11 |
awarded | Teacher |
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Jan 11 |
answered | The roots in a finite field |
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Dec 19 |
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Can we prove the completeness of FOL based on forcing? It may be worth mentioning that the same construction (considering some more dense sets) can also be used to prove the omitting types theorem. This may help explaining why it only works for countable theories, since unlike the completeness theorem, the omitting types theorem does not hold for uncountable theories. |
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Dec 19 |
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Can we prove the completeness of FOL based on forcing? As written, the construction does not produce a Henkin theory. $P$ should really be the poset of deductively closed consistent theories extending $T$ which are finitely axiomatized over $T$. Then for every formula $\phi(x)$, the set of conditions forcing $\exists x\,\phi(x)\to\phi(c)$ for some constant $c$ is also dense (due to the countable supply of fresh constants), hence the generic filter will give a Henkin theory. |
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Oct 30 |
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remainder of separable metrizable space @Brian: Sorry I didn’t notice this earlier, but in your answer, “second-countable $T_4$” and “separable metrizable” are the same thing, one is not more general than the other. |
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Oct 22 |
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remainder of separable metrizable space @ege: A metrizable space is separable iff it is second countable, and second countability is hereditary. |
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Apr 5 |
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How far do known ordinal notations span? You can effectively compare ordinals encoded using the same computable well-ordering. You cannot effectively compare two ordinals represented by codes of computable well-orderings, that would be tantamount to making $\omega_1^{CK}$ a computable ordinal, and it would also contradict Rice's theorem. |
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Mar 2 |
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Self-avoiding walk on $\mathbb{Z}$ @DejanGovc: Wrong. The set of such sequences is a Borel (in fact, $G_\delta$) subset of a Polish space, hence it is either countable or of cardinality $2^{\aleph_0}$. |
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Jan 17 |
awarded | Supporter |