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    $\begingroup$ The proof of the Baire Category Theorem I have in mind is a fairly direct generalisation of Cantor's Diagonal Argument. $\endgroup$ Commented Nov 22, 2010 at 18:03
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    $\begingroup$ Similarly, we can use the existence of (countably additive) Lebesgue measure to conclude that $\mathbb{R}$ is uncountable. Or on $2^{\mathbb{N}}$ the existence of the (countably additive) product measure. Or, from probability theory, the existence of an i.i.d. sequence of non-trivial random variables. $\endgroup$ Commented Nov 22, 2010 at 18:23
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    $\begingroup$ If you unwind the Baire Category proof of uncountability, it actually turns into a diagonal argument after all. Given any sequence $(x_i)$ of real numbers, you argue that the intersection over $i$ of $\mathbb{R} \setminus \{x_i\}$ is dense, by BCT, hence non-empty. The standard proof of BCT constructs an element of this intersection by first taking a ball contained in $\mathbb{R} \setminus \{x_0\}$, then a sub-ball of this contained $\mathbb{R} \setminus \{x_1\}$, and so-on. In other words, we construct a real by a countable sequence of approximations, [cont’d] $\endgroup$ Commented Nov 22, 2010 at 18:24
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    $\begingroup$ with the $n$th approximation ensuring that the resulting limit is not equal to $x_n$. So this is a version diagonal argument in the same sense that the nested limit theorem is. $\endgroup$ Commented Nov 22, 2010 at 18:25
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    $\begingroup$ @Todd: The abstract Baire category argument requires (dependent) choice, but for the reals can be done without requiring extra choice. Instead, enumerate the rationals, then when asked to "choose" a point in a given open interval, choose the first rational in that interval. $\endgroup$ Commented Nov 22, 2010 at 18:26