main blog @dronebetter (this is where i am creepy and weird) to be referred to as drone or tally hominids DNI i am running a social experiment... if you are a refugee (Palestinian or otherwise) who would like me to share your plea for money, please acknowledge this sentence; otherwise I will consider it proof that you are a bot account
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okay okay what about this... so get this one right... okay are you ready
"Damn girl did the seam along the underside of your penis get its request to fly rejected by TSA in the post-9/11 era??? because it is erm... Darkened in Pigmentation!"
troy's "sack" never felt quite the same after that day. also people kept looking at her funny. was it something about her new horsegirl friend? whenever asked, she would just giggle and comb troy's nascent mane and handfeed her a sugarcube. and she did not mind that at all, troy thought to herself, braying softly.
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math challenge: color the cartesian plane (RxR) with a countable infinity of colors such that no three distinct points with coordinates (a, b), (a, c), (d, b) all have the same color, or prove that this is impossible
Solution under cut. There may be a simpler way to show this.
The answer: neither of these is possible in general, because:
RxR can be colored by countably infinitely many colors in such a way iff CH holds.
(=>) Assume !CH, consider a subset of R of size ω2; it suffices now to show that ω2 x ω2 cannot be colored this way. For contradiction take a coloring C: ω2 x ω2 -> ω so that it is never true that C(a,b) = C(a,c) = C(d,b).
We now make excessive use of facts about stationary sets, namely:
if k is regular, points of any specific infinite regular cofinality below k are stationary
partitioning a stationary subset of a regular cardinal k into
more generally, Fodor's lemma: if S is stationary, and f:S->k is such that f(a) < a for all a in S, then f is constant on a stationary set.
Observe {α in ω2 | cof(α) = ω1} is a stationary set. Because there are only countably many colors, for every α in this set, there exists at least one n_α so that C(α,β) = n_α for unboundedly many β in α. Of course, there are countably many possible n_α, so we can find a fixed n and stationary S = {α in ω2 | cof(α) = ω1 and the β < α so that C(α,β) = n is unbounded}.
We take the function g : S -> ω2 which maps α to the minimal β<α so that C(α,β) = n. g takes some value ξ on a stationary subset T of S.
Now let α, α' in T. It follows C(α,ξ) = C(α',ξ) = n, and there is β where ξ<β<α so that C(α,β) = n.
(<=) Assume CH. It suffices to color ω1 x ω1 this way. For each a in ω1, pick an injection f_a from a to ω (each a is countable, so we can do this; presumably we are working in a ZFC setting). We color (a,b) by "diag" if a = b and otherwise by coloring with the pair (f_max(a,b)(min(a,b)), 1 iff max(a,b) = a and 0 otherwise). Obviously there are countably many colors.
Now, suppose any triangle (a,b), (a,c), (d,b) had the same color. Note it is impossible for any of these to lie on the diagonal, because then the other two would have to as well. Furthermore, either the first coordinate is always larger than the second, or the second coordinate is always larger than the first. But in the first case, f_a(b) != f_a(c); in the second case, f_b(a) != f_b(d). QED.
(Big thanks to @timestoppertactics; I worked out the second part of this first and then we figured out the first part together after I mentioned the problem to them)
you get to bond with your friend who administers them (who gets to see your abdomen and pinch a little part) and it quickly transpires to be less painful than you anticipated (unless they accidentally nicked a capillary in which case oh dear)
and then you spend the next two days with a slightly raised region that is sensitive to being bumped into anything at all and then la la la no taking tablets for an entire week
and also everything you experience causes you to cry (more than with pills) and your acne seemingly clears itself up immediately
People who don't actually know shit about math and go around saying that "Some infinities are bigger than others" do irreparable damage to my braincells
This is fair. I've always found myself not fond about that phrasing, admittedly. It's always felt a bit pop math (derogatory)-coded. I think this is partly to do with the invariable misapplication of this to wildly irrelevant notions of infinity, or more broadly to the lack of understanding of how many utterly unrelated notions of infinity there are in mathematics.
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misogyny might be one of the most brutal systems of oppression in the world and responsible for untold suffering over thousands of years but it also inspired a bunch of pathetic british guys with whiny voices to write some really good indie rock songs so its impossible to say if its bad or not
this is only a flop post because you don't know who Tim Misny is
Tim Misny is our local Saul Goodman style ~personal injury lawyer~ who's become kind of a local fixture because of his billboards. They started pretty straightforward:
But he became well known enough locally that he started doing this:
And finally, devolved into this:
If you wanted a modern TJ Eckleburg you COULD NOT DO BETTER than this man's fucking Dreamworks Eyebrow
surely an accidents lawyer would understand the majority of accidents are not clear-cut cases and so would desire to maximise the number of accidents to get a larger sample size from which to sift out good ones, making "please do not drink and drive" counterproductive to his Goals
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