Calculus TA

Problem Is the set of all irrational real numbers countable? Answer No. Suppose that the set of all irrational real numbers is countable. Then the set of real number is countable since it is a union of two countable sets, which is contradiction.

Problem A complex number $z$ is said to be algebraic if there are integers $a_0,\cdots,a_n$, not all zero, such that $$ a_0z^n + a_1z^{n-1} + \cdots + a_{n-1}z + a_n =0. $$ Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$ there are only finitely many equations with $$ n+|a_0| + |a_1| + \cdots + |a_n| = N.$$ Answer In Rudin's book, the definition of "countable" includes "infinity". Since the set of algebraic numbers contain all integers, it is an infinite set. It remains to show that it is at most countable. Define a set $A_N$ as follows: $$ A_N = \{p(z) = a_0z^n + a_1z^{n-1} + \cdots + a_{n-1}z + a_n \; | \; n + |a_0| + |a_1| + \cdots + |a_n| = N, a_i \in \mathbb{Z}, \text{not all zero}, n \in \mathbb{N}\}.$$ Then $A_N$ is a finite set. For each $p \in A_N$, let $B_p$ be a set of zeros of $p$, which is finite. Hence, the set $C_N:=\bigcup_{p \in A_N} B_p$ is also finite. Note that the set of algebraic number is $$ \b...

Answer Define a function $f:(0,1) \rightarrow (0,1]$ as $$f(x) = \begin{cases} 2x, & \text{$x = 1/2^n$ for $n=1,2,\cdots$} \\ x, & \text{otherwise} \end{cases}$$ (1) Injective: suppose $f(x_1) = f(x_2)$. If $f(x_1)=f(x_2)=1/2^n$ for some $n \in \{0,1,2,\cdots\}$, then $x_1=1/2^{n+1}=x_2$. Otherwise, $x_1=f(x_1)=f(x_2)=x_2$. (2) Surjective: suppose $y \in (0,1]$. If $y = 1/2^n$ for some $n \in \{0,1,2,\cdots\}$, then $y = f(1/2^{n+1})$. Otherwise, $y = f(y)$.

Problem Let $A$ be a set and let $\mathcal{P}(A)$ denote the set of all subsets of $A$. Prove that $A$ and $\mathcal{P}(A)$ do not have the same cardinality. Answer Suppose there is a bijection $f:A \rightarrow \mathcal{P}(A)$ and define $B=\{x \in A | x \notin f(x)\}$. Since $f$ is surjective, there exists $y \in A$ such that $f(y)=B$. If $y \in B$ then $y \notin f(y) = B$, which is contradiction. Similarly, if $y \notin B=f(y)$, then $y \in B$ by the definition of $B$. In either case, we get a contradiction.

Problem Let $f:A \rightarrow B$ be a function, $C_1,C_2 \subset B$, and $D_1,D_2 \subset A$. Prove (a) $f^{-1}(C_1 \cup C_2) = f^{-1}(C_2) \cup f^{-1}(C_2)$ (b) $f(D_1 \cup D_2) = f(D_1) \cup f(D_2)$ (c) $f^{-1}(C_1 \cap C_2) = f^{-1}(C_1) \cap f^{-1}(C_2)$ (d) $f(D_1 \cap D_2) \subset f(D_1) \cap f(D_2)$ Answer (a) $x \in f^{-1}(C_1 \cup C_2)$  $\Leftrightarrow$  there exists $y \in C_1 \cup C_2$ such that $y = f(x)$  $\Leftrightarrow$  there exists $y_1 \in C_1$ such that $y_1 = f(x)$ or there exists $y_2 \in C_2$ such that $y_2 = f(x)$  $\Leftrightarrow$  $x \in f^{-1}(C_1)$ or $x \in f^{-1}(C_2)$  $\Leftrightarrow$  $x \in f^{-1}(C_1) \cup f^{-1}(C_2)$ (b) $y \in f(D_1 \cup D_2)$  $\Leftrightarrow$  $y=f(x)$ for some $x \in D_1 \cup D_2$  $\Leftrightarrow$  $y=f(x_1)$ for some $x_1 \in D_1$ or $y=f(x_2)$ for some $x_2 \in D_2$  $\Leftrightarrow$  $y \in f(D_1)$ or $y \in f(D_2)$  $\Leftrightarrow$...