Ex 8: (in Python, using a reversal function)

def f(s):
return s[::-1]

dlmt = '"""'
code = """def f(s):
return s[::-1]

dlmt = '{}'
code = {}{}{}
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))"""
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))

Ex 4

Given a computable function $\varphi :S\mapsto \text{Comp}(S,\{T,F\left\}\right)$, define a function $f:S\mapsto \{T,F\}$ by the rule $f\left(s\right)=\{\begin{array}{cc}T& \varphi \left(s\right)\left(s\right)=F\\ F& \varphi \left(s\right)\left(s\right)=T\end{array}$. Then $f$ is computable, however, $f\notin \varphi \left(S\right)$ because for $s\in S$, we have that $\varphi \left(s\right)\left(s\right)=F⟹f\left(s\right)=T⟹\varphi \left(s\right)\ne f$ and $\varphi \left(s\right)\left(s\right)=T⟹f\left(s\right)=F⟹\varphi \left(s\right)\ne f$.

Ex 5:

We show the contrapositive: given a function halt, we construct a surjective function $f$ from $S$ to $\text{Comp}(S,\{T,F\left\}\right)$ as follows: enumerate all turing machines, such that each corresponds to a string. Given a $t\in S$, if $t$ does not decode to a turing machine, set $f\left(t\right)\equiv F$. If it does, let $\left[t\right]$ denote that turning machine. Let $f\left(t\right)$ with input $s\in S$ first run halt$(t,s)$; if halt returns $F$, put out $F$, otherwise $\left[t\right]$ will halt on input $s$; run $\left[t\right]$ on $s$ and put out the result.

Given a computable function $g:S\mapsto \{T,F\}$, there is a string $t^{\ast }\in S$ such that $\left[t^{\ast }\right]$ implements $g$ (if the turing thesis is true). Then $g=f\left(t^{\ast }\right)$, so that $f$ is surjective.

Ex 6:

Let $h:R\mapsto S$ be a parametrization of the circle given by $h\left(r\right)\mapsto \left(\cos \right(r),\sin (r\left)\right)$. Given $p\in S$ and $r\in R$, write $p+r$ to denote the point $h(p^{\prime }+r)$, where $p^{\prime }\in h^{-1}\left(\right\{p\left\}\right)$. First, note that, regardless of the topology on $X$, it holds true that if $f:X\mapsto S$ is continuous, then so is $f_r:x\mapsto f\left(x\right)+r$ for any $r\in R$, because given a basis element $\left(h\right(a),h(b\left)\right)$ of the circle, we have $f_r^{-1}\left(B\right)=f^{-1}\left(h\right(a)-r,h(b)-r)$ which is open because $f$ is continuous.

Let $\varphi$ be a continuous function from $X$ to $\text{Cont}(X,S)$. Then ${\varphi }^{\prime }:X\times X\mapsto S$ is continuous, and so is the diagonal function $h:x\mapsto \varphi (x,x)$. Fix any $r\in (0,2\pi )$, then $f:X\mapsto S$ given by $f:x\mapsto \left({\varphi }^{\prime }\right(x,x)+r))$ is also continuous, but given any $x\in X$, one has $\varphi \left(x\right)\left(x\right)={\varphi }^{\prime }(x,x)\ne {\varphi }^{\prime }(x,x)+r=f\left(x\right)$ and thus $\varphi \left(x\right)\ne f$. It follows that $\varphi$ is not surjective.

Ex 7:

I did it in java. There's probably easier ways to do this, especially in other languages, but it still works. It was incredibly fun to do. My basic idea was to have a loop iterate 2 times, the first time printing the program, the second time printing the printing command. Escaping the " characters is the biggest problem, the main idea here was to have a string q that equals " in the first iteration and " + q + " in the second. That second string (as part of the code in an expression where a string is printed) will print itself in the console output. Code:

package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?""+(char)34:"";String q=""+(char)34;q=i==1?q+"+q+"+q:q;String e=i==1?o+"+e);}}}":"System.out.print(o+";System.out.print(o+"package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?"+q+""+q+"+(char)34:"+q+""+q+";String q="+q+""+q+"+(char)34;q=i==1?q+"+q+"+q+"+q+"+q:q;String e=i==1?o+"+q+"+e);}}}"+q+":"+q+"System.out.print(o+"+q+";"+e);}}}

#1

Let f be such a surjection. Construct a member of P(S) outside f's image by differing from each f(x) in whether it contains x.

#2

A nonempty set has functions without a fixed point iff it has at least two elements. It suffices to show that there is no surjection from S to S -> 2, which is analogous to #1.

#3

T has only one element. Use that one.

#7 Haskell

source = "main = putStrLn (\"source = \" ++ show source ++ \"\\n\" ++ source)"
main = putStrLn ("source = " ++ show source ++ "\n" ++ source)

Is #8 supposed to read "Write a program that takes a function f taking a string as input as input, and produces its output by applying f to its source code. For example, if f reverses the given string, then the program should outputs its source code backwards."?

If so, here.

source = "onme = putStrLn $ f $ \"source = \" ++ show source ++ \"\\n\" ++ source"
onme f = putStrLn $ f $ "source = " ++ show source ++ "\n" ++ source

Ex 1

Exercise 1: Let $f:S\mapsto P\left(S\right)$ and let $A=\{x\in S|x\notin f(x\left)\right\}$. Suppose that $A\in f\left(S\right)$, then let $x\in S$ be an element such that $f\left(x\right)=A$. Then $x\notin f\left(x\right)=A⟹x\in A$ by definition, and $x\in A=f\left(x\right)⟹x\notin A$. So $x\notin A⟺x\in A$, a contradiction. Hence $A\notin f\left(S\right)$, so that $f$ is not surjective.

Ex 2

Exercise 2: Since $T$ is nonempty, it contains at least one element $t_0$. Let $h:T\mapsto T$ be a function without a fixed point, then $t_0\ne h\left(t_0\right)=:t_1$, so that $t_0$ and $t_1$ are two different elements in $T$ (this is the only thing we shall use the function $h$ for).

Let $\Psi :S\mapsto T^S$ for $S$ nonempty. Suppose by contradiction that $\Psi$ is surjective. Define a map $f:S\mapsto P\left(S\right)$ by the rule $f\left(x\right)=\{s\in S|\Psi (x\left)\right(s)=t_0\}$. Given any subset $U\subset S$, let $g:S\mapsto T$ be given by $g:x\mapsto \{\begin{array}{cc}{t}_0& x\in U\\ t_1& x\notin U\end{array}$ Since $\Psi$ is surjective, we find a $s^{\ast }\in S$ such that $\Psi \left(s^{\ast }\right)=g$. Then $f\left(s^{\ast }\right)=U$. This proves that $f$ is surjective, which contradicts the result from (a).

Ex 3

Exercise 3: By (2) we know that $T=\left\{x\right\}$, and so $f=\left\{\right(x,x\left)\right\}$ and $g=\left\{\right(s,h)|s\in S\}$ where $h=\left\{\right(s,x)|s\in S\}$. That means $x=g\left(s\right)\left(s^{\prime }\right)=f^n\left(g\right(s\left)\right(s^{\prime }\left)\right)$ for any $s,s^{\prime }\in S$. and $n\in N$.

Q7 (Python):

Y = lambda s: eval(s)(s)
Y('lambda s: print("Y = lambda s: eval(s)(s)\\nY({s!r})")')

Q8 (Python):

Not sure about the interpretation of this one. Here's a way to have it work for any fixed (python function) f:

f = 'lambda s: "\\n".join(s.splitlines()[::-1])'

go = 'lambda s: print(eval(f)(eval(s)(s)))'

eval(go)('lambda src: f"f = {f!r}\\ngo = {go!r}\\neval(go)({src!r})"')

I am confused by the introductory statement for #9. Is this an accurate rephrasing?

By representing syntax using arithmetic, it is possible to define a function $f$ as follows:

Define $f(s\in S_1,t\in S_1)$ with image in $S_0$, such that:
$f$ substitutes the Goedel-number of $t$ into $s$ (creating $s^{\prime }$) and then substitutes the Goedel-number of $s^{\prime }$ into some fixed formula to get a result in $S_0$.