other exercises: [[20241031_DiskMat_Serie6.pdf]]

Question for TA

Can you please correct all the tasks in this series? I have found the problems very difficult and I am not sure if my approach is correct

6.6 The hunt for Red October

inspiration: https://cage.ugent.be/~ldnaux/problem_5.html

The position of the submarine is

s(t)=s0+vt

Since s,v0Z are not known, the submarine has a par of unknown values (v,s0)Z×Z. Since Z is countable, Z×Z is countable as well (Theorem 3.22). The pairs can be written as

{(v1,s0,1),(v2,s0,2),,(vn,s0,n)}

where t=1,2,,n. At every t, Svetlana can fire a torpedo at

s(t)=s0,t+vtt

until she targets the position (pair) where the submarine is, e.g. t=n, thus sinking the submarine in a finite time.

6.5 Countability

For all l1,lN:

Al={f:N{0,1}|i=0kf(i)kl+1 for all kN}

~~Since k,lN, kl+1 cannot be smaller than 1. Thus, all functions

f:N{0,1}|i=0kf(i)1 for all kN

are also part of Al.~~
Now, we assume that Al is countable (AlN). Let

b{0,1}andb=b0b1b2g(b)=fbfb:N{0,1}fb(x)={1bx=1 and i=1x1bxx1l+10otherwise

Let b{0,1} and x,k,lN. Since the function only returns 1 if i=1x1bxx1l+1, the sum i=0kfb(i) is always smaller than kl+1. Thus, fb is in Al.

-> This means that g(b)=fbS for b{1,0}

Let b,b{0,1} so that bb. This means that bnbn for nN.
Let j be the smallest index where bjbj.
Then fb(j)fb(j), since only bj or bk satisfy the first case of the function.

-> fb is an injective function

We have shown that there is an injective function g:{0,1}Al. According to Definition 3.42, then applies {0,1}Al.

Since we assume that Al is countable and we have shown that {0,1}Al it follows (Lemma 3.15):

{0,1}AlAlN{0,1}N

This is a contradiction, thus proving that Al is uncountable.