All blackboards are gone. They were replaced during the break. (So, no chance to use this MathOverflow question on the near future.) Makes me think of T.H. Huxley’s On a Piece of Chalk (see here) and wonder what the equivalent will be in a few decades.

On the plus side, we now have computers and projection equipment on each classroom. I am using this quite a bit in my abstract algebra class. Except that, during the first few weeks, it was more often than not that the keyboard would be locked away.

Annoyed, I called OIT and asked that they please make sure it was unlocked before my class. It worked for a few days. But then, again, I found it locked.

I called again (I was charming, I am sure). So, somebody came to the classroom, looked at me, smiled. And pressed a button, to open the drawer.

Sigh.

(In my defense, the class is at 8:30 in the morning, and I’m supposed to drink less coffee these days. But still.)

(“Thanks. I’m sorry. I’m an idiot.” “Oh, no, no. It is new.” “It is a button.”)

This entry was posted on Sunday, February 19th, 2012 at 4:26 pm 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.

Have you had a chance to use a tablet+projection to replace the whiteboard? Infinite paper with infinite zoom to add additional comments in the right place is a great tool I use for taking notes, but never had a chance to use in class.

Let me address the first question. We can have models of $\mathsf{PA}$, $M\subsetneq N$ with $M$ cofinal in $N$. In fact, $M$ and $N$ do not even need to have the same cardinality. However, one can prove from the Davis-Matiyasevich-Putnam-Robinson theorem (on Hilbert's tenth problem) that already the assumption $M\subseteq N$ implies that $M$ is $\Sigma […]

The usual proof (as in Kanamori's book, section 11) is as follows: Work in $\mathsf{ZF}$. Note first, with Bernstein, that if $\omega_1\le\mathfrak c$, then there is a set of reals without the perfect set property: Either $\omega_1=\mathfrak c$, so $\mathbb R$ can be well-ordered, and we can build Bernstein sets using the usual transfinite recursion, or […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

The notions of Shelah cardinals and Woodin cardinals were introduced by Shelah and Woodin in their joint paper Large cardinals imply that every reasonably definable set of reals is Lebesgue measurable. Israel J. Math., 70 (3), (1990), 381–394. MR1074499 (92m:03087), which itself was the result of the hugely influential Martin's Maximum paper by Foreman, […]

The method of forcing certainly fits here. Before, set theorists expected that independence results would be obtained by building non-standard, ill-founded models, and model theoretic methods would be key to achieve this. Cohen's method begins with a transitive model and builds another transitive one, and the construction is very different from all the […]

This follows immediately from two facts: If $f$ is Riemann integrable, it is continuous almost everywhere. (This is part of Lebesgue's characterization of Riemann integrability: A function is Riemann integrable iff it is both continuous almost everywhere, and bounded. See for instance Theorem 7.6.5 in Abbott's Understanding analysis.) At any point […]

The set of all countable subsets of $\mathbb R$ has size $\mathbb R$. This is because there is an injection from this set into the set $\mathbb R^{\mathbb N}$ of countable sequences. Now, $\mathbb R$ has the same size as $\{0,1\}^{\mathbb N}$ (for instance, see here), and we see that $|\mathbb R^{\mathbb N}|=|(\{0,1\}^{\mathbb N})^{\mathbb N}|=|\{0,1\}^{\mat […]

Good question. The answer is no. Many functions are derivatives, though: For example, any continuous function. The fundamental theorem of calculus gives us that if $f$ is continuous, and $F$ is defined by $$ F(x)=\int_0^x f(t)\,dt, $$ then $f$ is the derivative of $F$. However, derivatives do not need to be continuous. But they are not arbitrary. For instanc […]

OK, we get $2^\kappa$ orders, as explained here. The answer does not change if we restrict our attention to scattered orders: Given any element of $(\kappa^+)^{

RT @replicakill: ...now I'm back in possession of those books, which I hope to pass on to my nephew. 5 days ago

RT @replicakill: ...When I left for college, I gave the collection to my sister who had become an avid reader by then... 5 days ago

RT @replicakill: ...I stacked them all in the center of my room and looked at them from every angle. Then I started reading them and didn't… 5 days ago

This is quite hilarious (I sympathize).

Have you had a chance to use a tablet+projection to replace the whiteboard? Infinite paper with infinite zoom to add additional comments in the right place is a great tool I use for taking notes, but never had a chance to use in class.

I haven’t, but I probably should. I’ve heard Hugh Woodin uses this quite effectively teaching precalculus, of all things.