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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..

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This is valid, provided you replace $\mathbb R^1\in\mathbb R^*$ by $\mathbb R^1\subset\mathbb R^*$. – Did Sep 10 '12 at 6:43
I see, that is clearer. – James Sep 10 '12 at 6:44
Not clearer, true instead of false. – Did Sep 10 '12 at 6:45
Is it because $\mathbb{R}^1$ is not an element? – James Sep 10 '12 at 6:47
$\mathbb{R}^1$ is not an element of $\mathbb{R}^*$ as it is not finite. But it is a subset. – Henry Sep 10 '12 at 6:49
up vote 1 down vote accepted

$\textbf{Hint}$ 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.

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That proves the bullet "$\mathbb{R}$ is uncountable $\rightarrow \mathbb{R}^1$ is uncountable" but I think James already knew that. – Henry Sep 10 '12 at 6:46

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