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\newtheorem{exa}{Example}
\begin{document}
\title{\textbf{Exercise}}
\author{Wang-Shiuan Pahngerei}
\date{April 8, 2012}
\maketitle

\begin{exa}
The set $\{ \varnothing , \{\varnothing\}  \}\cup \{   \{ \varnothing , \{\varnothing\}  \}  \}$ contains only 3 elements. So it has $2^3$ subsets. 
\end{exa}
\begin{exa}
Two sets $X$ and $Y$ are called identified, denoted by $X=Y$, provided that for each $x\in X$, $x\in Y$, and for each $y\in Y$, $y\in X$. Use it to show that whenever $\{a,b\} = \{c,d\}$,  we must have either $a=c, b=d$, or, $a=d, b=c$. 
\end{exa}
\begin{proof} 
Divid the statement according to whether $a=b$ and $c=d$, into $4$ cases, we might step by step get the conclusion.  
\end{proof}
\begin{exa}
$A=B$ if and only if $\wp A = \wp B$. 
\end{exa}
\begin{proof}
Assume that $A=B$, we want to show $\wp A=\wp B$. Let $K\in \wp A$. Then $K\subset A$. Since $A\subset B$, yet we have $K\subset A\subset B$. We get $K\subset B$, i.e. $K\in \wp B$. Hence $\wp A\subset \wp B$. Similarly we have $\wp B \subset \wp A$. Therefore $\wp A= \wp B$. 

Conversely, assume that $\wp A = \wp B$, we ought to show that $A=B$. Let $x\in A$. Then $\{x\}\subset A$, and we obtain $\{x\}\in \wp A$. Because $\wp A = \wp B$, $\{x\}\in \wp B$. Hence $\{x\}\subset B$, so we get that $x\in B$. Hence $A\subset B$. Similarly we have $B\subset A$. Therefore $A=B$. 
\end{proof}


\begin{exa}
$T\times (\cup_{a\in A} S_a)=\cup_{a\in A} T\times S_a$. 
\end{exa}
\begin{proof}
Let $x\in T\times (\cup_{a\in A} S_a)$. Then $x=(t,s)$ for some $t\in T, s\in\cup_{a\in A} S_a$. Accordingly we know that $s\in S_{a_0}$ for some $a_0\in A$. Hence the ordered pair $(t,s)$ is such that $t\in T$ and $s\in S_{a_0}$, and then $(t,s)\in T\times S_{a_0}$. Then we find that $x\in\cup_{a\in A} T\times S_a$. Hence $T\times (\cup_{a\in A} S_a)\subset\cup_{a\in A} T\times S_a$.

Conversely, given $y\in\cup_{a\in A} T\times S_a$, then $y\in T\times S_{a_0}$ for some $a_0\in A$. We can write $y=(t,s)$ for some $t\in T$ and $s\in S_{a_0}$. It follows that $s\in \cup_{a\in A} S_a$, and $y=(t,s)$ is in fact in $T\times  \cup_{a\in A} S_a$. Hence $T\times (\cup_{a\in A} S_a)\supset\cup_{a\in A} T\times S_a$. Therefore, $T\times (\cup_{a\in A} S_a)=\cup_{a\in A} T\times S_a$.
\end{proof}

\begin{exa}
$\cap_{j=1}^\infty [0,\frac{1}{j}) = \{0\}$.
\end{exa}
\begin{proof}
Let $x\in \cap_{j=1}^\infty [0,\frac{1}{j})$. We ought to show that $x\in \{0\}$. i.e. $x=0$. Assume not. If $x\textless 0$, then $x\notin [0,\frac{1}{1})$, which is not permitted. Else if $x\textgreater 0$, then we choose an $N\in\mathbb{N}$ such that $\frac{1}{N}\textless x$. Since $x$ is in the intersection, $x$ must be in $[0,\frac{1}{N})$, a contradiction. Therefore, $x=0$, and $\cap_{j=1}^\infty [0,\frac{1}{j}) \subset \{0\}$.

Conversely, given $y\in \{0\}$, then $y=0$. Since we know that $0\leq 0\textless \frac{1}{j}$ for each $j\in\mathbb{N}$, $y\in\cap_{j=1}^\infty [0,\frac{1}{j})$. Hence $\cap_{j=1}^\infty [0,\frac{1}{j}) \supset \{0\}$. Therefore $\cap_{j=1}^\infty [0,\frac{1}{j}) = \{0\}$.
\end{proof}

\begin{exa}
Let $c_n$ be a strictly decreasing sequence of positive real numbers. If $\lim_{n\to\infty}c_n =0$ then $\cap_{j=1}^\infty [0,c_n) = \{0\}$. 
\end{exa}
\begin{proof}
Let $x\in \cap_{j=1}^\infty [0,c_n)$. We have to show that $x=0$. It is clear that $x\geq 0$. It remains to show that $x\textgreater 0$ would never occur. Assume that $x\textgreater 0$, then because of the limit of the sequence there is an $N\in\mathbb{N}$ such that for each $n\geq N$, we have $c_n \textless x$. For this $N$, we can observe that $x\textgreater c_N$, but according to the intersection, $x\in [0,c_N)$, a contradiction. Hence $x=0$. 

The converse direction follows from the last example, which is not hard to prove. Hence $\cap_{j=1}^\infty [0,c_n) = \{0\}$.
\end{proof}

\begin{exa}
The way to negate a statement is to exchange the quantifiers, and then negate the pattern forms. If we hope to negate that "for each $\varepsilon\textgreater 0$, there is an $N\in\mathbb{N}$, such that for every $n\in\mathbb{N}$, if $n\textgreater N$ then $|a_n - a|\textless\varepsilon$", we will write down the result: "There is an $\varepsilon_0\textgreater 0$ such that for each $N\in\mathbb{N}$, there is an $n_N\in\mathbb{N}$, such that $n_N\textgreater N$ but $|a_n - a|\geq\varepsilon_0$ ".
\end{exa}

\begin{exa}
A way to express an ordered pair by means of an unordered pair is that: we may define $\langle a,b\rangle = \{\{a\}, \{a,b\}\}$. We ought to show that if $\langle a,b\rangle = \langle c,d\rangle$, then $a=c$ and $b=d$. 
\end{exa}
\begin{proof}
By $\textit{example\,\,3}$ we find that either (I) $\{a\}= \{c\}$ and $\{a,b\}=\{c,d\}$ or (II)  $\{a\}= \{c,d\}$ and $\{a,b\}=\{c\}$.

In case (II) we obtain that $c=d=a$ and $c=d=b$ because we know that $c\in\{c,d\}$ and $c$ will consequencely belong to $\{a\}$, so that  $a=c$ and similarly $a=d$. The later case is alike. Hence $a=c=b=d$. 

In case (I), we know by $\{a\}= \{c\}$ that $a=c$. we firstly assume that $a=b$, then similarly we have $a=c=b=d$. If $a\ne b$, we hope to show that $c\ne d$. Aussume the contrary, that $c=d$, then the same case that $a=b=c=d$ would again occur, a contradiction. So $c\ne d$. Since $b\in \{a,b\}$, $b\in \{c,d\}$. If $b=c$ then $a=b=c$, which is excluded, so $b=d$. 
\end{proof}



\begin{exa}
Let $x\geq 0$, $n\in\mathbb{N}$. Show that 
\begin{align}
[x] + [x+\frac{1}{n}]+\cdots + [x+\frac{n-1}{n}]=[nx]\notag
\end{align}
where [x] denotes the Gaussian Notation. 
\end{exa}
\begin{proof}
Define $I_k = [\frac{k-1}{n}, \frac{k}{n})$. Then for each $x\geq 0$, there is a $k$ such that $x\in I_k$. We hope to perform induction on $k$. i.e. When $x\in I_k$, the equality holds. 
\begin{itemize}
\item[(1)] If $k=1$, then $0\leq x\textless \frac{1}{n}$. We find that $0\leq x+\frac{j}{n}\textless x+\frac{n-1}{n} = 1$ for $j=0,1,2,\cdots ,n-1$. So that $[x+\frac{j}{n}]=0$. On the other hand, since $0\leq nx \textless n\cdot \frac{1}{n} =1$, $[nx] = 0$. Hence the equality holds. 
\item[(2)] Assume that when $x_0 \in I_k$, $[x_0]+\cdots + [x_0 +\frac{n-1}{n}] = [nx_0]$. Let $x\in I_{k+1}$. Then we know that $x-\frac{1}{n}\in I_k$. By induction hypothesis we obtain
  \begin{align}
    [x-\frac{1}{n}] + [x] +\cdots + [x-\frac{1}{n} +\frac{n-1}{n}] &= [n\cdot (n-\frac{1}{n})]\notag\\
      &=[nx-1]\notag\\
      &=[nx]-1\notag
  \end{align}
  So that
  \begin{align}
    [x] +\cdots + [x-\frac{1}{n} +\frac{n-1}{n}]+[x+\frac{n-1}{n}] &= [nx]-1+[x+\frac{n-1}{n}] -[x-\frac{1}{n}] \notag\\
        &=[nx]-1\,\,\, + \,\,\,1\notag\\
        &=[nx]\notag
  \end{align}
\item[(3)] By M.I, the equality holds for each natural number $k$, and then it holds no matter which inteval $I_k$ $x$ belongs to. Hence the equality holds for all $x\geq 0$. 
\end{itemize}
\end{proof}


\end{document}