Wednesday, February 09, 2011

An exercise in visualisation

Many people think that the only neighbourhood of a dense subset of \(X\) must be the entire space \(X\). There's a counter-example that's so easy it's all one can think of afterwards, so that one soon forgets how hard it is to visualise: let \(q : \mathbb Z_{> 0} \to \mathbb Q\) be an enumeration of \(\mathbb Q\), and consider \(U_1 := \bigcup_{j \in \mathbb Z_{> 0}} (q_j - 1/2^j, q_j + 1/2^j)\). Clearly, \(U_1\) is a neighbourhood of the dense subset \(\mathbb Q\) of \(\mathbb R\). (Actually, this is two counterexamples for the price of one, since \(\mathbb Q\) is also co-dense.) The Lebesgue measure of \(U_1\) is then at most \(\sum_{j = 1}^\infty 1/2^{j - 1} = 2\); in particular, it's not all of \(\mathbb R\).

I forget where I first saw this. It is too easy to think that you are picturing \(U_1\), by envisioning a collection of blobs; so try visualising its complement instead. This is a closed subset of \(\mathbb R\), of infinite measure, that contains no intervals. This is maybe not so scary if you're used to visualising, or imagining that you've visualised, the Cantor set, but it's still pretty strange.

That's not the visualisation exercise. Instead, I ask you to visualise the following, drawn from Oxtoby's book. Notice that the construction of \(U_1\) can easily be generalised to give, for any \(\varepsilon > 0\), a neighbourhood \(U_\varepsilon\) of \(\mathbb Q\) of measure at most \(2\varepsilon\). Now consider \(G := \bigcap_{i = 0}^\infty U_{1/2^i}\). This is a neighbourhood of \(\mathbb Q\) whose ‘biggest radius’ \(1/2^i\) shrinks to \(0\), so it must be just \(\mathbb Q\), right? No. Notice that each \(\complement U_{1/2^i}\) is the complement of an open, dense set in \(\mathbb R\), hence is (closed and) nowhere dense; so \(\complement G = \bigcup_{i = 0}^\infty \complement U_{1/2^i}\) is of the first category in \(\mathbb R\). If we had \(G = \mathbb Q\), then \(\mathbb R = \mathbb Q \cup \complement G\) would again be of the first category; but Baire assures us that it is not.

It is easy to see where the reasoning goes wrong—since the intervals that go into the construction of each \(U\) are badly overlapping, the ‘largest radius’ need not actually be shrinking to \(0\) as we expect—but not, or at least not for me, easy to get any idea what \(G\) actually looks like. Thoughts?

UPDATE 2011-02-09: Note, incidentally, that \(G\) is also a Lebesgue-null set. It is obviously uncountable (or else the above argument would go through essentially unchanged), but it must be a very small sort of uncountable. I wonder if assuming that it has the cardinality of the continuum is equivalent to the Continuum Hypothesis?


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