It’s a number that is greater than every number of the form (\frac{9}{10^n}, n \in {1, 2, 3, \dotso}) but less than (1).
Note that my claim is that this is precisely the meaning of the symbol (0.\dot9). It represents the infinite sum (\sum_{n=1}^{\infty} \frac{9}{10^n}) which in turn represents a number that is greater than every number of the form (\frac{9}{10^n}, n \in {1, 2, 3, \dotso}) but less than (1).
Since it’s less than (1), it’s clear that it’s not a number larger than every integer. But it’s also not a number that can be represented as a finite sum of rational numbers.
But my point is precisely that rearranging the remaining points in the line does not show that the number of points did not change.
We started with a set (A = {P_1, P_2, P_3, P_4, \dotso}). We removed every odd point to get a new set (A’ = {P_2, P_4, \dotso}). Since we preserved the identity of points, we can see that (P_1) and (P_3) are missing from (A’). Note that we didn’t merely change the position of points. If that were the case, (A’) would include (P_1) and (P_2) (as well as all other odd points.) Also note that we’re actively disregarding where on the line these points are. (P_1) does not necessarily represent the first point on the line. (P_1) can be ANY point on the line. The only condition is that it’s the only point on the line that is represented by this symbol.
Let’s go back to your example.
We have two infinitely long lines (A) and (B) each represented by a set of points that constitute them.
(A = {a1, a2, a3, \dotso})
(B = {b1, b2, b3, \dotso})
(a1) represents the first point on the line (A), (a2) the second point on the line (A), and so on.
Similarly, (b1) represents the first point on the line (B), (b2) the second point on the line (B), and so on.
Then, we pick any point on the line (A) and rename it to (p1). Let’s take (a3) and rename it to (p1).
(A = {p1, a1, a2, a4, \dotso})
Then, we do the same for (B). Let’s pick (b1) and rename it to (p1).
(B = {p1, b2, b2, b3, \dotso})
We then repeat this process for all other elements. But before we can do this, we have to specify how big these sets are in relation to each other. Are they equal in size or is one of them larger than the other? In your example, you said that the two parallel lines are equally long, so that means we have to declare that the two sets are equal in size. This gives us the following result:
(A = {p1, p2, p3, \dotso})
(B = {p1, p2, p3, \dotso})
This allows us to say that there is a one-to-one correspondence between the two sets based on the simple observation that every element in (A) is also present in (B) and vice versa. (We couldn’t do this with what we had at the start.)
Now, we take set (B) and take every second point out. What do we get? We get:
(B’ = {p1, p3, p5, p7, \dotso})
We can now compare (B’) to (A) and conclude that (A) is greater than (B’) because every element in (B’) is present in (A) but not every element in (A) is present in (B).