How can I prove that set T is countably infinite? Countable and uncountable sets; rational numbers Let S be the collection of all non-vertical lines in the 2-dimensional plane $R^2$ passing through the origin. We can index the collection $S$ using $R$ as the index set, as follows.
For each $i∈R$, define $$L_i={\{(x,y) ∈ R^2 | y = ix}\}$$
Note that $i$ is simply the slope of the line $L_i$. We can then write $$S={\{L_i | i∈R}\}$$

*

*What is $⋃_{i∈R} L_i$ ?

*Explain why $S$ is uncountable.

*We shall call a line $L_i∈_S$ special if at least one point on the line $L_i$ other than the origin has rational numbers for both coordinates (i.e., there is at least one point $(x,y)≠(0,0)$ on $L_i$ such that $x∈Q$ and $y∈Q$). Let $T=\{L_i\in S\ |\ L_i\text{ is special}\}$. Prove that $T$ is countably infinite.


$R$ denotes set of real numbers and $Q$ denotes set of rational numbers.

my answers:

*

*$⋃_{i∈R} L_i$ is the union of all the lines with slope $i \in R$.

*$S$ is uncountable because the set $R$ of real numbers is uncountable since $S$ is indexed using $R$ as the index set.

*I am not sure how to prove this. What i have in mind is, since $T$ is a special set with a special line that has at least one point with rational numbers for both its coordinates, should I show that $T$ is countable because the set of rational numbers $Q$ is countable? Please help.

Also are my answers to 1 and 2 correct? I somewhat feel they are but would love to hear from you all.
 A: *

*Your answer doesn't make sense. The set $\bigcup_{i\in\Bbb R}L_i$ is a subset of $\Bbb R^2$ and you should say which set this is. It's $\Bbb R^2\setminus\{(0,y)\mid y\in\Bbb R\setminus\{0\}\}$.

*By that argument, if $L_i=\{(0,0)\}$, then the set $\{L_i\mid i\in\Bbb R\}$ would be uncountable too, but, it fact, it consists of a single element. Your set is uncountable because, for each $i\in\Bbb R$, $(1,i)\in L_i$, but $(1,i)\notin L_j$ when $j\ne i$. So, when $i\ne j$, $L_i\ne L_j$, and therefore the map $i\mapsto L_i$ is injective. So, since $\Bbb R$ is uncountable, the set os all $L_i$'s is uncountable too.

*The set $T$ is countable because $\Bbb Q^2\setminus\{(0,0)\}$ is countable. So, only countably many lines can contain points of $\Bbb Q^2\setminus\{(0,0)\}$.

A: Yeah, your solutions are (almost) correct. However, the notion of being "indexed" using an index set is not clearly defined. It should be defined as follows:
A set is $S$ indexed by $R$ if there is a bijective mapping between the index set $R$ and the set $S$, i.e. for $i\neq j\in R$ we have $L_i\neq L_j\in S$. The important point is the condition of $i\neq j$ implying $L_i\neq L_j$ because otherwise, one can just take the set of one line $\{L\}$ and assign this line to all real numbers. In that case we have also a mapping $\mathbb{R} \to \{L\}$, but that is not bijective and certainly $\{L\}$ is not uncountable infinite.

*

*It is clear that the union of those lines is the union of all lines, that's basically the definition. Therefore I would argue that $$\bigcup_{i\in\mathbb{R}} L_i = \mathbb{R}^2 \setminus (\{0\} \times \mathbb{R}\setminus \{0\}) = (\mathbb{R}\setminus\{0\} \times \mathbb{R}) \cup \{(0,0)\}$$ as the union of those lines cover almost full $\mathbb{R}^2$ and only the $x$-axis (without $(0,0)$) is not included in that union.

*See above paragraph about the indexing of those lines.

*You can answer the third question similar to the second. If there is at least one rational point on a line, that point has a rational slope. Therefore, $T$ can be indexed by $\mathbb{Q}$ and is indeed countably infinite.

A: Your answer 1. and 2. are indeed correct.
Regarding 3., you're also right. $T$ is infinite countable because $\mathbb Q$ is infinite countable and because when a line passing through the origin also passes through a point having both rational coordinates, then the slope of such a line is also rational.
A: 1.Union of such lines gives $R^2$ -{Y axis} .(0,0) is included.
Since Y- axis is vertical is not included in it. For any other point in $R^2$ We can find a line passing through it and origin.
2.Yeah. Its uncountable. Since positive Reals are uncountable and for any +ve real number we can find a corresponding line with that slope. Thus a bijection map between postive reals and Positive sloped lines(through origin) .
3.A line having rational numbers for both coordinates other than origin say $(r_1,r_2)$ will have a rational slope $\frac{r_1}{r_2}$. Thus we can create a bijection between rationals and such lines. So they are countable.
