I Think "high Value Person" Can Be Interpreted As "someone Who Is Liked By Everyone And Perfectly Adheres

in a way. over the last two years or so. mathematics has become the altar at which I pour out my private grief, and transmute it to something like solace. it does not particularly matter to me if I am ever any good at it. what matters is that the effort I apply to it is rewarded by understanding. I have no natural aptitude for it; I am climbing this hill because it was the steepest and least hospitable to me. there is less agony in the gentler slope, but less valor

may I add some very red trains

red inside as well

If anyone wondered what do I mean when I'm saying our trains are "very green":

And yep inside they are also green:

(Usually not that nice looking though)

The ones in my area usually look like this ↑

(all pics are from Google images lol)

Master Control Program

A minimal 74 knot on the simple cubic lattice

(source code)

http://proof.ucalgaryblogs.ca/

This is the best resource for studying math that I've found in a while! It's 300+ pages of flawed/incorrect proofs on topics including logic, analysis, and linear algebra. Each flawed proof is followed by a classification of its errors, and a corrected version.

uhh probably the worst math feeling is when you're so excited about proving something and you talk about it to someone who does math with you and they say oh but it's trivial

I'm reblogging this to compare it later with 1.A from Hatcher's Algebraic Topology. in that chapter he defines the topology on a graph if anyone else wants to check it out

Intuitively, it seems to me that graphs should be some sort of finite topological space. I mean, topology studies "how spaces are connected to themselves", and a graph represents a finite space of points with all the internal connections mapped out. That sounds topological to me! And of course many people consider the Seven Bridges of Königsberg problem to be the "beginning" of topology, and that's a graph theory problem. So graphs should be topological spaces.

Now, I vaguely remember searching for this before and finding out that they aren't, but I decided to investigate for myself. After a bit of thought, it turns out that graphs can't be topological spaces while preserving properties that we would intuitively want. Here's (at least one of the reasons) why:

We want to put some topology on the vertices of our graph such that graph-theoretic properties and topological properties line up—of particular relevance here, we want graph-theoretic connectedness to line up with topological connectedness. But consider the following pair of graphs on four vertices:

On the left is the co-paw graph, and on the right is the cycle graph C_4.

Graph theoretically, the co-paw graph has two connected components, and C_4 has only one. Now consider the subgraph {A, D} of the co-paw graph. Graph theoretically, it is disconnected, and if we want it to also be topologically disconnected, it must by definition be the union of two disjoint open sets. Therefore, in whatever topology we put on this graph, {A} and {D} must be open. The same argument shows that {B} and {C} must be open as well. Therefore the topology on the co-paw graph must be the discrete topology.

Now consider the subgraph {B, D} of C_4. It is disconnected, so again {B} and {D} must be open. Since {A, C} is also disconnected, {A} and {C} must be open. So the topology on C_4 must again be the discrete topology.

But these graphs aren't isomorphic! So they definitely shouldn't have the same topology.

It is therefore impossible to put a topology on the points of a graph such that its graph-theoretic properties line up with its topological properties.

Kind of disappointing TBH.

yea right in some parallel universe

All that I understand about algebraic geometry at my present stage of learning.

The proof is left as an exercise to the IRS

My favourite fucked up math fact™ is the Sharkovskii theorem:

For any continuous function f: [a,b] -> [a,b], if there exists a periodic point of order 3 (i.e. f(f(f(x))) = x for some x in [a,b] and not f(x) = x or f²(x) = x), then there exists a periodic point of ANY order n.¹

Yes you read that right. If you can find a point of order 3 then you can be sure that there is a point of order 4, 5, or even 142857 in your interval. The assumption is so innocent but I cannot understate how ridiculous the result is.²

For a (relatively) self-contained proof, see this document (this downloads a pdf).

(footnotes under read more)

¹ The interval does not have to be closed, but it should be connected. (a,b), (a,b] and [a,b) all work.

² Technically the result is even stronger! The natural numbers admit a certain ordering called the Sharkovskii ordering which starts with the odd primes 3 > 5 > 7 > ... , then doubles of primes, then quadruples of primes and so forth until you get no more primes left, ending the ordering in 2³ > 2² > 2. Sharkovskii's theorem actually says that if you have a periodic point of order k, then you have periodic points of any order less than k in the Sharkovskii ordering. It is frankly ridiculous how somehow prime numbers make their way into this mess.