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\begin{document}
\title{\textbf{Where to go ??}}
\author{\textbf{The Essense of Schr$\ddot{\text{o}}$der-Bernstein Theorem}}
\date{2016/12/28}
\maketitle

A significant question is that: Whether when $A\prec B$ and $B\prec A$, it is necessary that $A\approx B$ ??  
Before handling this, we firstly consider a practical example. 

\begin{exa} To consider whether  $[0,1]\approx [0,1)$\footnote{This is changed from previous version that there is only one added point}, we ask if there's a bijective $f:[0,1]\to [0,1)$. Suppose it exists, who is the images of $1$? It may be chosen, say $f(1)=\frac{1}{2}$, and then, since $f$ should be injective, who is $f(\frac{1}{2})$ ? If we set for example that $f(\frac{1}{2})=\frac{1}{4}$, we still have define $f(\frac{1}{4}),f(\frac{1}{8})$ and so on. Now, the idea emerges. 

Define 
\begin{align}
  f(x)=
    \begin{cases}
      \frac{x}{2},  &\text{if $x\in\{\frac{1}{2^k}\,:\,k=0,1,2,\cdots \}  $;}\\
      x,   &\text{otherwise.}
    \end{cases}
\end{align}
We will confirm ourselves their equinumerousity. 
\end{exa}

\begin{exa}[Schr$\ddot{\text{o}}$der-Bernstein Theorem] 
If $A\prec B$ and $B\prec A$, then $A\approx B$.

Draw two lines, by which we represent $A$ and $B$. Then pick up an injection $f$ from $A$ onto $f(A)\subset B$ (We assume $f(A)\ne B$). The key point is that we will pretend $f(A)$ and $A$ are the same. (At least we say they are closely related, indistinguishable, or essentially the same.) 

Now we only focus on $B$ because $A$ is viewed as a subset of $B$. Let 
\begin{align}
B_0 = B\setminus f(A)\notag.
\end{align}

The one-to-one function $h:B\to A$ is now viewed from $B$ onto the ``subset" $A$, as in the previous example that $\frac{x}{2}$ sends $[0,1]$ injectively into $[0,1)$. The where to go question occurs on the function $h$ and we are looking for it. So we want to answer where $B_0$(In Example 1 case, it is $=[0,1]\setminus[0,1)=\{1\}$) is sent,  and who plays a role as $x/2$ in Example 1? 

But don't forget we already have such an one-to-one function $h:B\to A$ by condition, so everything is well-prepared and we should focus on a  precise proof. From now on, since we are going to operate $f$ and $h$, we should write down the embedding function $f:A\to B$ when we ``iterally apply" $h$ on $B_0$. 

 Define
\begin{align}
B_1&=f(h(B_0))\notag\\
B_2&=f(h(B_1))\notag\\
&\vdots\notag
\end{align}
Then the following is a bijection $g$ from $B$ onto $A$. Choose

\begin{align}
  g(x)=
    \begin{cases}
      h(x),  &\text{if $x\in B_j $;}\\
      f^{-1}(x),   &\text{otherwise.}
    \end{cases}
\end{align}
Then all above is the same as Example 1. 
\end{exa}

\begin{exa}
This is another theorem about the method: If $X\supset Y\supset X_1$, and $X\approx X_1$, then $X\approx Y$. 

The way we solve it is like this: Since we have an $f$, one-to-one from $X$ onto $X_1$, yet $f(Y)\subset X_1$ (Assume that $f(Y)\ne X_1$), denoted by $Y_1$. It's clear that $f$ is one-to-one from $Y$ onto $Y_1$, so $X_2:=f(X_1)$ is a proper subset of $Y_1$. Continually repeat this process, we obtain the family of sets
\begin{align}
X=X_0\supset Y=Y_0\supset X_1\supset Y_1\supset X_2\supset Y_2\supset\cdots\notag
\end{align}

All we perform above are natural. We might now start our "$\textit{where-to-go}$" method. Eventually we find an $F$ from $X$ to $Y$ (Note that $F$ is $\textbf{from X to Y}$.), by 
\begin{align}
  F(x)=
    \begin{cases}
      f(x),  &\text{if $x\in X_j \setminus Y_j$;}\\
      x,   &\text{otherwise.}
    \end{cases}
\end{align}
Finally, we ought to show that $F(x)$ indeed works. 
\end{exa}

An important remark is: 
\begin{exa}
The last two examples are equivalent.
\end{exa}
\end{document}