Associate to each sequence $a=\{\alpha_n\}$, in which $\alpha_n$ is $0$ or $2$, the real number $$x(a)=\sum_{i=1}^{\infty}\frac{\alpha_n}{3^n}$$ Prove that the set of all $x(a)$ is precisely the standard Cantor set.
First of all I'd like to say that I know how to do it in the right way, but when my roomie shows me his solution, I think it is kind of problematic. I want to refute him but I do not know what argument to use :-|
His solution
Notations: $E$ is the standard Cantor set. $E_k$ is the $k-$th stage of removal during the construction of $E$, namely, $E_0:=[0,1],E_1=[0,\frac13]\cup[\frac23,1],E_2:=[0,\frac19]\cup[\frac29,\frac13]\cup[\frac23,\frac79]\cup[\frac89,1],\cdots$
Define a map $f$ that maps a sequence $(\alpha_1,\alpha_2,\cdots)$ into the real line as follows:
Imagine you land randomly upon $E_0:=[0,1]$, say, at $0.423$.
Now the removal construction (or rather, destruction :P) begins. You're now forced to find a safe place on an surviving interval. And our oracle $f$ will tell you what to do so that you can survive.
Before the first removal begins, $f$ shows you $\alpha_1$. If $\alpha_1=0$, go to the interval to you left, namely, $[0,\frac13]$ (If you are already here, stay put); if $\alpha_1=2$, go to or stay in the right, namely, $[\frac23,1]$.
So you survive the first removal, but the second one is coming in no time. So $f$ shows you $\alpha_2$. This time, similarly, go to or stay in the nearest left interval (from your current location, of course) if $\alpha_2=0$, and go to or stay in the right if $\alpha_2=2$.
The pattern is: before the $k-$th removal, the oracle $f$ tells you $\alpha_k$, but not $\alpha_{k+1},\alpha_{k+2},\cdots$, and you accordingly ($0$=left, $2$=right) decide which direction to go to take shelter.
For example, if the sequence is $(0,2,0,\cdots)$, your survival path will be $$[0,\frac13]\to [\frac29,\frac13]\to[\frac29,\frac7{27}]\to\cdots$$ Each $a$ characterizes an "infinite" survival path, and due to nested closed interval theorem each "infinite" survival path leads to exact one point in $E$. Therefore the oracle $f$ is indeed an injection from the set of all $a$s (a subset of $\Bbb R^{\infty}$) into $E$ (this paragraph is where I want to refute most!).
$\cdots$(Then my roomie intended to show $f$ is surjective, which I think is extremely unconvincing and is therefore omitted)
I think his argument doesn't hold water because it uses something like an "infinite" path (in his point of view the elements of Cantor set are just terminals of such paths), which I cannot accept as a well-defined mathematical object.
If we accept this "path-terminal" theory I think the proof for "nonempty perfect set in a metric space is uncountable" would be too simple: just take distinct $x_1$, $x_2$, then you can find disjoint neighbourhoods $N(x_1), N(x_2)$. Since $x_{1,2}$ are limit points, you can again find $x_{11,12}\in N(x_1)$ and $x_{21,22}\in N(x_2)$. And you can continue such progress forever, to obtain an "infinite binary forking tree" whose paths have a one to one correspondence with the set of all sequences consisting solely of $0,1$ and is hence uncountable. Because each path characterizes a "terminal", the terminals (a subset of the whole perfect set) are uncountable as well, so the perfect set in question must be uncountable. This "proof" is, if not totally wrong, at least much too flippant to be considered as mathematically rigorous, I think.
Or am I simply being fuzzy? Is the series $$x(a)=\sum_{i=1}^{\infty}\frac{\alpha_n}{3^n}$$ actually in the same spirits as something like an "infinitely long" path?