Bandidas

This movie was horrible. Not Soul plane horrible or even My super ex-girlfriend horrible, but I had to pause for a second to come up with those two examples. Unbelievable that Salma Hayek actually appears with a straight face in a small special in the DVD explaining how Penélope Cruz and herself wanted for so long to collaborate, to make this movie together. How insulting! Avoid it like the plague.

One Response to Bandidas

Harsh J says:

March 17, 2007 at 10:49 pm

Yeah it did suck but then, people do anything to watch Salma’s.

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Answer by Andrés E. Caicedo for Conflating reals and sets of countable ordinals "nicely" August 15, 2017

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

Andrés E. Caicedo

Answer by Andrés E. Caicedo for What is the need for MathSciNet reviewing again the journal versions of conference papers? August 6, 2017

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for product of power sets July 13, 2017

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Choosing *two* representatives July 19, 2017

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Axiom of Choice: Ultrafilter vs. Vitali set July 6, 2011

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

Andrés E. Caicedo

Math.StackExchange

Answer by Andrés E. Caicedo for Definition for "power" of a tree? (Kurepa trees) August 21, 2017

The power of a set is its cardinality. (As opposed to its power set, which is something else.) As you noticed in the comments, Kurepa trees are supposed to have countable levels, although just saying that a tree has size and height $\omega_1$ is not enough to conclude this, so the definition you quoted is incomplete as stated. Usually the convention is that […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Why does this proof that Every Well-ordered Set is Isomorphic to a Unique Ordinal need to use the Axiom of Replacement August 21, 2017

The key problem in the absence of the axiom of replacement is that there may be well-ordered sets $S$ that are too large in the sense that they are longer than any ordinal. In that case, the collection of ordinals isomorphic to an initial segment of $S$ would be the class of all ordinals, which is not a set. For example, with $\omega$ denoting as usual the f […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Which sets are Lebesgue measurable in $ZFC$? August 20, 2017

R. Solovay proved that the provably $\mathbf\Delta^1_2$ sets are Lebesgue measurable (and have the property of Baire). A set $A$ is provably $\mathbf\Delta^1_2$ iff there is a real $a$, a $\Sigma^1_2$ formula $\phi(x,y)$ and a $\Pi^1_2$ formula $\psi(x,y)$ such that $A=\{t\mid \phi(t,a)\}=\{t\mid\psi(t,a)\}$, and $\mathsf{ZFC}$ proves that $\phi$ and $\psi$ […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Reorder the terms of the series August 19, 2017

Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm: For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the ﬁrst $p$ positive terms, then the ﬁrst $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. Th […]

Andrés E. Caicedo

Answer by Andrés E. Caicedo for Are there uncountably infinite complete extensions of finite theories of the natural numbers? August 11, 2017

Yes, by the incompleteness theorem. An easy argument is to enumerate the sentences in the language of arithmetic. Assign to each node $\sigma $ of the tree $2^{

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Bandidas

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