The real line with doubled origin is weakly equivalent, but not homotopy equivalent, to \(S^1\).

Pf: Both spaces are the homotopy pushout of the diagram \(* \leftarrow \mathbb R-\{0\} \rightarrow *\), so they're weakly equivalent.

They aren't homotopy equivalent because any map out of the line with doubled origin to a Hausdorff space factors through the Hausdorffification, which is the contractible space \(\mathbb R\).

#proofinatoot

The only increasing continuous involution of \(\mathbb R\) is \(id\).

Pf: Let \(f\) be such an involution. Say there exists a point with \(f(x) \neq x\). Say \(x < f(x)\). Then since \(f\) increasing, \(f(x) < x = ff(x)\), a contradiction. If \(f(x) < x\), then \(ff(x) = x < f(x)\), again a contradiction. Thus \(f = id\).

If anyone else was wondering why on earth the homotopy groups of the motivic sphere spectrum are related to quadratic forms, I highly recommend this paper by Cazanave:

https://arxiv.org/pdf/0912.2227.pdf.

TLDR: Endomorphisms of \(\mathbb P^1\) correspond to rational functions, and there's a quadratic form called the Bezout form associated to any rational function. Naive \(\mathbb A^1\) homotopies are taken into account via stable isomorphism of quadratic forms. Group complete to get non-naive.

#proofinatoot

#jokes

My friend showed me this, and I'm not sure who to attribute it to originally.

\(\sqrt[3]{2}\) is irrational.

Pf: Assume for the sake of contradiction that \(\sqrt[3]{2} = \frac{p}{q}\), is rational. Cubing, we get \(2q^3 = p^3\), or \(q^3 + q^3 = p^3\). This is a contradiction of Fermat's Last Theorem.

It's a minor miracle when any counting argument I do ends up being correct.

Also, the dimension of the moduli space of Lagrangian subspaces of the hyperbolic quadratic space \(\mathbb H^n\) is \(\frac{n(n-1)}{2}\).

The statement can be found in the orthogonal grassmannian section of this paper by Eisenbud, Popescu, Walter: https://arxiv.org/pdf/math/9906171.pdf.

#proofinatoot

Let \(A:\mathbb R^n\rightarrow \mathbb R^n\) be a real matrix with positive entries. Then \(A\) has a positive eval with an evec whose entries are all positive.

Sketch: Consider the part of the unit sphere in the non-negative orthant. Defining \(f(v) = Av/|Av| \) we get a map from this subset to itself. Since this subset is homeomorphic to \( D^{n-1} \), the Brouwer fixed pt thm tells us it has a fixed point \(x\). Thus \(x = Ax/|Ax|\), so \(Ax=|Ax|x\), and we're done.

#proofinatoot

For every \(n \geq 1\) there exists an uncountable subset of \(\mathbb R^n\) s.t. any \(n\) distinct elements in the subset are linearly independent.

Pf: If \(n=1\) take \(\mathbb R^n - \{0\}\). If \(n \geq 2\), consider the set of elements of form \( (1,t,t^2,\dots,t^{n-1})\) as \(t\) ranges over \(\mathbb R^n - \{0\}\). This is an uncountable set. To see that any \(n\) distinct elements are independent, put them into a matrix and note that it's a Vandermonde matrix.

Joined Jun 2017