# $[0,1]$ is uncountable using Baire's Category Theorem

Given that, $\mathbb{R}$ is complete.
$[0,1] \subset \mathbb{R}$ and is closed $\implies [0,1]$ is complete and clearly non-empty.
So we know by BCT: $[0,1]$ is second category.

Suppose $[0,1]$ is countable, then let $[0,1] = \cup_{n\in\mathbb{N}}\{x_n\}$
Singletons are closed and have empty interior $\implies$ nowhere dense.
So $[0,1]$ is first category by def *Contradiction.

Is this proof concrete enough?

• I don't know what you mean by "concrete enough", but the proof is solid. – Theo Bendit Aug 2 '18 at 3:59
• Solid enough :) – Devendra Singh Rana Aug 2 '18 at 4:08
• – Kenny Lau Aug 2 '18 at 4:30