## (Working on) a non-algebraic proof that S_n is not contractible

For a blog which tries to add something to philosophy as well, its pauses and unfinished threads strike me as less desirable…sorry! I haven’t forgotten about the threads, but other more pressing matters seem to crop up continually.

In the meantime I think I found a nice non-algebraic proof that S_n is not contractible. The proof only uses some continuous-function theory, specifically the Michael Selection Theorem. I’m trying to expand this to a non-algebraic non-simplicial proof of the general Jordan Curve theorem. The first part I will write down somewhere next month, I hope.

Update: unfortunately I made a mistake in my proof-sketch, which I discovered when (finally) writing the proof down. After trying to fix it, I discovered a counterexample to one of its essential arguments. Since mistakes can be a little illuminating also, let me indicate the trouble:

Trouble:
There is a (simple) continuous function $f$ from $[0,1]\times [0,1]$ to $[0,1]$ such that for all $x\in [0,1]$ we have $f(x,0)=0$ and $f(x,1)=1$ and yet there is no continuous function $t$ from $[0,1]$ to $[0,1]$ such that for all $x\in [0,1]$ we have $0.

Perhaps a nice challenge to find such a function $f$ yourself… it’s not really hard :-).

Spoiler: (don’t read if you want to find a solution yourself) Picture the square $[0,1]\times [0,1]$, in this square draw a bold S smack in the middle (the letter should have some breadth, say $b$). Elongate the  lower tip of the S toward a $b$-breadth interval around the point $(0, \frac{1}{2})$ and the upper tip toward a $b$-breadth interval around the point $(1, \frac{1}{2})$. Now let the continuous function $f$ be such that $f$ assumes the value $1$ on all the points in the square above the elongated S, and $0$ on all the points in the square below the elongated S, and a suitable gradient in between on the points of the elongated S itself.