Connectedness of R\QXQ

This one was just amazing.

Consider R\QXQ. Is it connected?

It turns out that it is polygonally path connected with at most two segments!!

And the proof is just two lines:


Consider two points p,q in R\QXQ.

Then, through p (and similarly for q) there is an infinity of lines that pass through only irrational points. Pick a line each for p and q. Call their intersection r. Then p->r->q  is the desired path.

Kolmogorrov's 0-1 Law

We learned the Kolmogorrov's 0-1 Law today in class.

And I had this (very very informal) intuition:

In the 0-1 Law we talk about the tail events, i.e., the events belonging to the tail sigma algebra. The tail sigma algebra is defined as

F_tail = \cap_{n >= 1}  \sigma(X_n, X_n+1, X_n+2, ... )

That is, it consists of events that are independent of finitely many initial X_n's. 

Obviously, F_tail \subset F := \sigma(X_1, X_2, ....), the "full" sigma algebra. 

In a way, to construct the F_tail, we 'drop' the initial X_n's one by one. For each X_n dropped, the sigma algebra becomes smaller but is still a sigma algebra. Note that this sigma algebra encodes lesser bits of information.

Thus, 'in the end', when you have 'dropped all the X's', you have a sigma algebra but you do not have any information bits left - and that sigma algebra is effectively the trivial sigma algebra. Thus any event in it must have probability 0 or 1!

Of course, this is very very vague and there is a LOT of handwaving in the arguments - but I thought the intuition was cool.

As a corollary, it would also be interesting to consider the effective information content of a probability measure on a sigma algebra. That is, in the tail sigma algebra above, the effective information content is 1 bit.