How to prove that set T is countably infinite? Countable and uncountable sets; rational numbers

discrete mathematicselementary-set-theoryrational 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 $$S={\{L_i | i∈R}\}$$

  1. What is $⋃_{i∈R} L_i$ ?
  2. Explain why $S$ is uncountable.
  3. 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:

  1. $⋃_{i∈R} L_i$ is the union of all the lines with slope $i \in R$.
  2. $S$ is uncountable because the set $R$ of real numbers is uncountable since $S$ is indexed using $R$ as the index set.
  3. 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.

Best Answer

  1. 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 $\Bbb R^2\setminus\{(0,y)\mid y\in\Bbb R\setminus\{0\}\}$.
  2. 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.
  3. The set $T$ is countable because $\Bbb Q^2\setminus\{(0,0)\}$ is countable. So, only countably many lines can contain points of $\Bbb Q^2\setminus\{(0,0)\}$.
Related Question