Consider the injections and given by
, where or .
(a) From the proof of Cantor-Schroder-Bernstein compute F(n) directly for n =
0, 1, 2, 3, 4 and find .
Is this just simply finding g o f? Like for n=2 it would be f(2) = 2 then g(0,2) = 6?
Or is it... Find x such that (g o f)(x) = 0,1,2,3,4?
OR! Is it something to do with the fact that... if i=0 then F(n) can be either 0,3,6, ... (3n + 0 basically). And if i=1 then F(n) can be either 1,4,7, ... (3n+1)..?
Since F isn't actually defined I'm not sure what it is.
AND. Since the CSB theorem is involved, there will be bijection A -> B so it makes me think that some values of n will equal 0,3,6, ... or 1,4,7 ... and the rest will have to be defined as something else...
Bleh im dumb.
Proof is the one with the zig-zig diagram that has A on one side, B on the other, f is an injection from A to B and B is an injection from B to A.
Then you repeatedly apply f and g to an element .
Now, we must be in one of two situations: either
(a) we can go up indefinitely from a or end on the A side, or,
(b) we end on the B side.
Let , a is type a, , a is type b. Then and .
From this we have a well defined map given by...
if , if
Taken from the notes!
Indeed I probably shoulda noticed this bit first...
So, Does this mean that F(0), F(1), etc = f(0), f(1), ... respectively since they are single integers and since its of the form (i,n), so we apply g to it..?