# Applications of Baire's Threom [duplicate]

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In a lecture on Baire's Theorem (for complete metric spaces), I gave, for a rather advanced undergraduate class in Real Analysis (covering the theory of metric spaces and elements of general topology), I realised that all the applications I provided were either straight-forward or could be proved using more elementary tools. For example:

a. $\mathbb Q$ is not complete.

b. $\mathbb R$ is uncountable. (Otherwise, if $\mathbb R=\{x_n\}_{n\in\mathbb N}$, then $\bigcap_{n\in\mathbb N} (\mathbb R\setminus\{x_n\})$ would be dense and empty.)

c. A complete metric, with the property that every point is an accumulation point, is uncountable.

Could you suggest more interesting applications? They do not have to be too easy to prove.

## marked as duplicate by Nate Eldredge, Andrés E. Caicedo, Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 23 '14 at 15:32

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• "Almost all" continuous functions are nowhere differentiable... This is one of Munkres' applications. – Henno Brandsma Oct 23 '14 at 15:19

## 2 Answers

In the course I learned this in, one of the examples (with a sketch proof) was that there exists a continuous function that is nowhere differentiable, as in Henno's comment.

In a different course, we proved this by constructing an explicit example, which was certainly much more complicated (even if more elementary) than fleshing out the sketch proof.

We also proved that the Cantor set is uncountable (and some slightly odd properties of continuous functions), but I haven't thought seriously about whether these are easy to prove in other ways.

It can be used to (easily) show that the rational numbers $\mathbb{Q}$ are not a $\mathrm{G}_\delta$-subset of $\mathbb{R}$. This in turn can be used to show that there does not exist a function $f: \mathbb{R} \to \mathbb{R}$ which has exactly the set of rational numbers as its points of continuity!