1
$\begingroup$

Show all finite subsets of reals is uncountably infinite (or is it?).

Firstly, I assumed that "all finite subsets of reals" is equivalent to the Kleene closure of $\mathbb{R}$, $$\mathbb{R}^* = \mathbb{R}^0\cup\mathbb{R}^1\cup\mathbb{R}^2\cup...$$

  • $\mathbb{R}$ is uncountable. $\Rightarrow \mathbb{R}^1$ is uncountable.
  • $\mathbb{R}^1 \subset \mathbb{R}^* \Rightarrow \mathbb{R}^*$ is uncountable because the union of an uncountable set with another set is also uncountable.
  • $\mathbb{R}^*$ is uncountably infinite.

Is this a valid proof? I am sort of new to the subject of proofs..

  • 1
    This is valid, provided you replace $\mathbb R^1\in\mathbb R^*$ by $\mathbb R^1\subset\mathbb R^*$.2012-09-10
  • 0
    I see, that is clearer.2012-09-10
  • 2
    Not clearer, true instead of false.2012-09-10
  • 0
    Is it because $\mathbb{R}^1$ is not an element?2012-09-10
  • 0
    $\mathbb{R}^1$ is not an element of $\mathbb{R}^*$ as it is not finite. But it is a subset.2012-09-10
  • 0
    In general, $A\cup B=C\implies A\subset C$.2012-09-10

1 Answers 1

1

$\textbf{Hint}$ Singletons are finite subsets. How many singleton subsets of $\mathbb{R}$ are there?


The Kleene closure is a way of computing all the finite subsets but introducing it is not exactly necessary since the main idea really is $\mathbb{R}^1 \subset \mathbb{R}^*$, which is exactly the hint.

  • 0
    That proves the bullet "$\mathbb{R}$ is uncountable $\rightarrow \mathbb{R}^1$ is uncountable" but I think James already knew that.2012-09-10