There is this cemetery near our house and, near the cemetery, there is a playground Francisco wanted to visit.

Off we go, crossing the cemetery. Najuma and Francisco are ahead of me, he busy picking up pine cones, she pulling the wagon that Francisco insisted we take with us. I am distracted, reading the last few pages of a homework set I want to finish grading.

There is a wooden sort of plank on the ground. You see where this is going, don’t you?

Anyway, I did not even get to step on the thing, wafer-thin as it was. As I am putting my foot down, I realize the mistake. Gravity is a marvelous thing; it is the first thing I tried to teach Francisco, 3 years or so ago, how gravity works; this is just a practical lesson. Because the wood breaks and falls and I fall with it, 6 feet or so. Najuma said later that she heard a loud noise, a rumbling coming from the ground. She turned, and saw me disappear.

The two of them approach as I try to get up, and somehow manage to find enough strength to lift myself out of the grave. (As a kid, this is the sort of thing I never had enough strength to do. The one time I went rock climbing, it was what stopped me from reaching higher than I did.) Francisco thinks it is a game and wants to go down the hole as well. Crying ensues.

And that’s it, really. Somehow I managed not to injure myself. Except for a bit of dirt here and there, you cannot really tell anything happened (the pages I was reading are somehow neatly stacked next to the hole). We looked in vain for somebody, so they can cover the hole.

“Good thing you are not superstitious,” Najuma tells me.

43.614000-116.202000

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Sunday, November 7th, 2010 at 7:45 am and is filed under Life. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

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

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 […]

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 […]

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. […]

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 […]

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 […]

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 […]

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 $\psi$ […]

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 […]

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^{