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 $$\mathbb R$$.

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:

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.

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: arxiv.org/pdf/math/9906171.pdf.

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.

In my proofinatoot, I just noticed that there's a minor typo; t should range over (\ \mathbb R - \{0\} \),

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.

Testing $$\LaTeX$$.

$$x^n + y^n = z^n$$.

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