This is ~~not~~ totally a bookmark.

# Infamous Heel-Filcher

## Monday, November 24, 2014

## Wednesday, November 19, 2014

### Metal for winter

So, there's these Russians, and they have a band...

... and it makes me think of snow and ice and other things nice...

... and it makes me think of snow and ice and other things nice...

## Wednesday, October 8, 2014

### Blut Aus Nord

The two tracks currently available for streaming in promo for Blut Aus Nord's new album are... fucking amazing.

That is all.

That is all.

## Monday, October 6, 2014

### If you don't have student loans, there's a high probability you're missing something

An opinion piece floated across my radar this morning; penned by one Jessica Slizewski, originally at XOJane and picked up at Time.com, it announces to the reader that I Don’t Have Student Loans and I Don’t Feel Bad for People Who Do. OK, so I'm being trolled. I don't care, I feel the need to retort.

Before I being, I do feel compelled to state that, like Ms Slizewski, I strongly feel that the

## Thursday, September 25, 2014

## Sunday, September 14, 2014

### A universal algebraist proves Fodor's Lemma

In the last post, we mentioned Fodor's Lemma (in the context of an attempted proof that didn't end up working out). Well, my brain was idling the other evening, and decided that it needed to prove this lemma. Don't ask my why my brain does what it does.

Attention conservation notice: this post is written for someone with maybe a lower-level grad-school level of knowledge, or maybe upper-level undergrad, and who knows what a subalgebra is but doesn't necessarily know any logic.

Attention conservation notice: this post is written for someone with maybe a lower-level grad-school level of knowledge, or maybe upper-level undergrad, and who knows what a subalgebra is but doesn't necessarily know any logic.

## Monday, September 8, 2014

### More on the countable chain condition

We last met the ccc in the context of preserving cofinalities in forcing extensions; but the exercise that I gave the proof for was specifically about forcing. This week, I ran across a nice little exercise which doesn't explicitly mention forcing at all -- it's pure combinatorics -- but, at least for me, thinking about it using the forcing idea was the key to solving the problem.

At first glance, this looks completely obvious -- things are either compatible or incompatible, and you can't have more than countably many pairwise incompatible things. The problem, though, is that the compatibility relation need not be transitive -- just because you have an uncountable subset of \(X\) which is not an antichain does

What made this problem interesting to me was that I had to

**Exercise 1:**Let \( \mathbb{P} \) be a ccc poset, and let \( X = \{ x_i \colon i < \omega_1 \} \subseteq \mathbb{P} \) be a subset. Show that there exists an uncountable subset of \( X \) whose every pair are compatible.At first glance, this looks completely obvious -- things are either compatible or incompatible, and you can't have more than countably many pairwise incompatible things. The problem, though, is that the compatibility relation need not be transitive -- just because you have an uncountable subset of \(X\) which is not an antichain does

*not*mean that it satisfies the conclusion of the exercise.What made this problem interesting to me was that I had to

*discard*some of the heuristics that usually serve me well. In particular, a dependable heuristic when dealing with posets is duality: if something is true for all posets, then you should be able to turn your poset upside-down and it should still be true. However, the compatibility relation is not stable under duality! Most forcing posets have a single weakest condition; if you dualize, you get that every two conditions are compatible, which is clearly useless.
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