You have the uncountable set $A$ and its countable subset $B$. You want to show that $|A\setminus B|=|A|$, i.e., that there is a bijection between $A$ and $A\setminus B$. To see what that would mean, let’s imagine for a moment that we already have such a bijection $h:A\to A\setminus B$. $A$ is the union of the disjoint sets $B$ and $A\setminus B$, and $h$ is a bijection, so $h[A]$, which is $A\setminus B$, is the union of the disjoint sets $h[B]$ and $h[A\setminus B]$, as in the sketch below. And because $h$ is a bijection, we know that $B$ and $h[B]$ have the same cardinality; in particular, $h[B]$ is countable. In other words, if we’re to have this bijection $h$, we need to have some countable subset $h[B]$ of $A\setminus B$ that $h$ matches up with the subset $B$ of $A$. The set $C$ of your proof is going to be that set.
We know that $A\setminus B$ has a countably infinite subset $C$, and we know from an earlier result that $B\cup C$ is countably infinite. This means that $|B\cup C|=|C|$: there is a bijection $f:B\cup C\to C$. We can now use $f$ to build the desired bijection $h:A\to A\setminus B$ quite easily. We’ll split $A$ into two pieces, $B\cup C$ and the rest, which is $A\setminus(B\cup C)=(A\setminus B)\setminus C$; these are shown in red and white, respectively, in the lefthand set in the picture below. We’ll use the bijection $f$ to map $B\cup C$ to $C$, and we’ll use the identity map $\mathrm{id}$ to send $(A\setminus B)\setminus C$ to itself. Combining these two bijections gives us a bijection, which I’ll call $h$, from $A$ to $A\setminus $B$, as in the picture below.
Formally we define $h:A\to A\setminus B$ by
$$h(a)=\begin{cases}
f(a),&\text{if }a\in B\cup C\\\\
a,&\text{if }a\in(A\setminus B)\setminus C\;.
\end{cases}$$
The key idea is simply that all countably infinite sets are the same size, so we can find a bijection between any two of them. We use that fact to find the bijection $f$ that ‘collapses’ $B\cup C$ into $C$; then we leave all of the other elements of $A$ alone (i.e., those in $(A\setminus B)\setminus C$), and the net effect is to collapse $A$ into $A\setminus B$ with the bijection $h$.
Yes it look's correct, basically what you construct is a piece of the space that has the same cardinality as $\mathbb(R)$ and that is only covered with countably number of points. On the same theme would be that each line touches the unit circle at at most two points, making those points where a line touches the unit circle countable, but the unit circle consist of uncountable points.
I would run into measure theory and realize that the integral of the characteristic function of a line (ie a function being one for every point the line crosses and zero every where else) is zero. Then so is the integral of the supremum of those function. That is the line misses almost every point in the plane.
Best Answer
You can List all lines in the plane this way:
Use the fact that a line is described uniquely once you know its slope, and one of its intercepts with the axes. The slope is indexed by the Reals, and so is the intercept, say the x-intercept. So we count all the possible pairs ( slope, x-intercept), and show it is equal to $|\mathbb R|$
1) Consider all lines thru the origin $(0,0)$. These are described uniquely by their slope, and there are $|\mathbb R|$ of them, since the slope is parametrized by the Reals.
2) From 1) , we can cover all other cases of lines not going thru $(0,0)$ , by considering all possible ( say x-) intercepts of a line thru any point, with fixed slope$m$. For every line in 1), there are $\mathbb R$ lines not going thru the origin, but with the same slope. This means there are $|\mathbb R|\times |\mathbb R|=|\mathbb R|$ total lines in the plane.