my question reads as follows:
I have constructed a proof and am concerned about 2 things:
1) The validity of my proof.
2) The construction of my proof.
I am asking for someone to read through it and tell me where I can improve my proof for presentation.
Proof:
The set $W$ is uncountable.
Assume $W$ is countable. By definition, $W$ is countable if there exists a function, $f:W\to\mathbb{N}$ , such that $f$ is a bijection. Furthermore, countability implies that the elements of $W$ can be arranged as a list. Suppose that all the possible elements of $W$ can be expressed as follows:
$\omega_{0} = x_{00},x_{01},x_{02},x_{03},x_{04},.....$
$\omega_{1} = x_{10},x_{11},x_{12},x_{13},x_{14},.....$
$\omega_{2} = x_{20},x_{21},x_{22},x_{23},x_{24},.....$ (1)
$\omega_{3} = x_{30},x_{31},x_{32},x_{33},x_{34},.....$
$. . . . . .$
,where $\omega_{i}\in W$ , $x_{ij}\in\{0,1\}$ for $i,j\in\mathbb{N}$ . In addition, the restriction on $W$ implies that
$x_{ij}+x_{i(j+1)}+x_{i(j+2)}\neq\begin{cases} 3 & ,\ as\ (x_{ij},x_{i(j+1)},x_{i(j+2)})=(1,1,1)\\ 0 & ,\ as\ (x_{ij},x_{i(j+1)},x_{i(j+2)})=(0,0,0) \end{cases}$, (2)
for all $x_{ij}\in\{0,1\}$ . Assuming $W$ is countable implies that our bijective function, $f$ , is defined as $f(w_{i})=i$ .
Let $\tau_{n}=x_{n0},x_{n1},x_{n2},x_{n3},x_{n4},\cdots$ for $n\in\mathbb{N}$ , be an infinite binary adhering to condition (2) . Let $x_{n0}=\{0,1\}\backslash\{x_{00}\}$ and $x_{n1}=\{0,1\}\backslash\{x_{11}\}$ . This implies that $\tau_{n}$ is distinct from binaries $\omega_{0}$ and $\omega_{1}$ . In determining the value of $x_{n2}$ , caution must be taken to not only create a distinction from $\omega_{2}$ but to also satisfy condition (2) . Thus we let
$x_{n2}=\begin{cases} \{0,1\}\backslash\{x_{22}\} & ,\ if\ x_{n0}+x_{n1}+\{0,1\}\backslash\{x_{22}\}\neq3\ or\ 0\\ x_{22} & ,\ if\ x_{n0}+x_{n1}+\{0,1\}\backslash\{x_{22}\}=3\ or\ 0 \end{cases}.$
If $x_{n0}+x_{n1}+x_{22}\neq3$ or $0$ , then $(2)$ has been satisfied and $\tau_{n}$ is distinct from $\omega_{3}$ . Thereafter, proceed to make a distinction from $\omega_{4}$ using the same procedure. $If x_{n0}+x_{n1}+x_{22}=3$ or $0$ , then while $\tau_{n}$ has satisfied $(2)$ it is not distinct from $\omega_{2}$ . Thus, continue to find a distinction from $\omega_{2}$ by iterating this process for $x_{23}$ .
That is,
$x_{n3}=\begin{cases} \{0,1\}\backslash\{x_{23}\} & ,\ if\ x_{n1}+x_{n2}+\{0,1\}\backslash\{x_{22}\}\neq3\ or\ 0\\ x_{23} & ,\ if\ x_{n1}+x_{n2}+\{0,1\}\backslash\{x_{22}\}=3\ or\ 0 \end{cases}.$
Continue this process until a distinction from $\omega_{2}$ is made. That is,
$x_{np}=\begin{cases} \{0,1\}\backslash\{x_{3p}\} & ,\ if\ x_{n(p-2)}+x_{n(p-1)}+\{0,1\}\backslash\{x_{3p}\}\neq3\ or\ 0\\ x_{3p} & ,\ if\ x_{n(p-2)}+x_{n(p-1)}+\{0,1\}\backslash\{x_{3p}\}=3\ or\ 0 \end{cases}$.
Once a distinction has been made proceed to make a distinction from $\omega_{3}$ following the same process above. Continuing this overall iteration creates a binary $b_{n}$ such that it is not contained in the list, $(1)$ . Therefore $W$ can not be listed and thus is uncountable. $\square$
Thank you all in advanced for you time !
Two suggestions: (a) Show that eventually, there will be a distinction. (b) Once you assume that $W$ is countable, you can go on to say that $W$ can be expressed as $\ldots$ (Also, feel free to use $w$ instead of $\omega$, but this doesn't matter.) On a similar note, giving a bijection $f: W \to \mathbb{N}$ is exactly the same thing as listing the elements of $W$.
However, if I wanted to prove your problem, I'd want to use one of the following:
Lemma 1 Assume $A$ is uncountable. If there is an injection $A \hookrightarrow B$, then $B$ also is uncountable.
or
Lemma 2 Assume $B$ is uncountable. If there is a surjection $A \to B$, then $A$ also is uncountable.
For lemma 2, $B$ should could be the collection of all sequences of 0's and 1's, which is uncountable. In your case, $W$ is $A$. For lemma 1, this is swapped, and applying lemma 1 should be easier then lemma 2.