Archive for helmikuu, 2020

Cartesian product of countable infinities is countably infinite proof

Suppose we have two countably infinite sets, which means we can find injection (one-to-one) relation to natural numbers. 

In other words, we could literally assign natural number indexes to only one of each member of the set.

So, we have with cardinality , and with cardinality

Then cartersian product (set of all possible pairs from A and from B) – is also infinite, but countable. I mean, for , we can assign single natural number for each member of P.

Lets prove it. Suppose we injections f and g, so that and

Lets define function this way:

Then for and , because of  unique division into primes and f,g injectivity (a,b)=(c,d).

Therefore there is injection from p(a,b) to natural numbers, and cartesian product is infinitely countable.

About number construction

– Let us construct natural numbers (0,1,2,3..)
– Then from naturals, lets construct integer numbers (…,-2,-1,0,1,2…)
– Then from integers, lets construct rationals (…,-0.3,…,0,…,1/2,..,0.75,..)
– And finally from rationals – real numbers (…,-0.3,…,0,…,1/π,…,1/2,..,0.75,..,e,…,π,…)
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Elementary Linear Cellular automata example

Lets consider linear cellular automata with Wolfram rule 90

Elementary cellular automata is essentially string of two symbols, lets call them 0 and 1, equipped with rules (and starting configuration, of course) 

Rule takes symbol in the position, as well as left and right neighbors. Ie – when calculating symbol of the cell, we should consider 3 cells, cell to the left, cell itself and cell to the right.

Ie – suppose we have set of rules(*):
111 - 0
110 - 1
101 - 0
100 - 1
011 - 1
010 - 0
001 - 1
000 - 0

Or graphically:


So if we have 10011 for starting configuration, calculation goes like this(from top to bottom):

1 0 0 1 1
(0) 1 1 1 (1)
(1) 1 0 0 (1)

(italics) means that we dont know which symbol is to the left and right to the starting configuration, for simplicity I consider here absent symbols to be 0s (ie …00000100110000…)

Now, what is 90, which is called Wolfram rule.

It is obvious, that  3 cells define state, and since we have two symbols, all possible combinations are , and since we have two possible symbols for each combination – rule of CA can be expressed as (0,2⁸=256) number.

90 in binary would be (*), which is enough to describe elementary ca behavior.

Lebesque measure of line on R² surface

There were question in exam, which I understood very close to correct, thus absolutely wrong.

Roughly translated: “Define real axis ℝ Lebesque outermeasure on ℝ² surface

Lets get into it. Bit intuitively first.. Suppose we have segment of a line (a,b)

segment (a,b) on real axis

segment (s) itself can be expressed as

Actually, lets forget about s, lets consider without segments, ie

Then outer measure can be expressed as union of open sets:

for all

Where intervals are:
Thus length of intervals are from one integer to another – ,.. all integers
Height of intervals are

Ie , where biggest height is , which is quater of epsilon.


or shorter: and for all

It is obvious from here

This article was inspired by Margarita. – reborn

I think I’d better come back to good old, instead of spamming facebook page.

Oh yea, it all seems to work now.. kinda enjoy new editor for WordPress. So – here are plans for the future:

  • Lebesque integral and measure, few notes regarding course and exam (english/finnish)
  • Cellular Automata (english)
  • Bitcoin shop with tor in mind (english)
  • Немножко политики/russian politics (russian/english)

There. Lebesque integral notes&measure is coming today.
CA – on this weekend, based on university lecture.
Bitshop – next week.
Russian politics – when feel like it.