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A closed meagre subset of $[0,1]$ is either countable or homeomorphic to the Cantor set: either way it is $0$-dimensional.

Q.1. Is every closed meagre subset of an $n$-dimensional locally compact Hausdorff space of dimension $\le n-1$ (for $n\ge 1$)?

Q.2. If so, is there anything unusual about meagre sets in $0$-dimensional and infinite-dimensional spaces in which closed meagre sets can have dimension equal to the dimension of the whole space (one would think that meagre sets might be 'fatter' in some sense in these cases)?

Q.3. Is there a simple example of an uncountable closed meagre subset of the Cantor set?

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3 Answers 3

up vote 4 down vote accepted

For Q3, the answer is yes. Think of the Cantor set as consisting of the points that have a base-3 expansion containing only 0's and 2's. Now take the subset of those points where the 2's occur only in even-numbered positions.

Also, in the first sentence, "either countable or homeomorphic to the Cantor set" isn't right; it could be the union of a countable set and a copy of the Cantor set. Nevertheless, it would still be 0-dimensional.

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Q1. The answer is NO.

Take $\mathbb R$ and attach to each rational point $p/q$ a segment of length $1/q$. You get a locally compact 1-dimensional metric space $M$. The set $\mathbb R$ forms a closed meagre subset in $M$.

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Thanks for these two answers (Andreas and Anton). I am sorry that I have only one tick to allot. Anton's example looks a useful one to remember. –  Douglas Somerset Apr 14 '12 at 7:53

By the way, your question 1 is true for n-dimensional manifolds (since it is true for $\mathbb{R}^n$ by theorem of Menger and Urysohn).

Note that Q3 can also be answered noting that Cantor set $C$ is homeomorphic to $C^2$ and for every point $x$, {$x$} $\times C$ is a meagre subset of $C^2$. Note that this construction gives you Andreas's answer if x=0.

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Thanks for this: it gives an easy way of visualizing a meagre subset of $C$. –  Douglas Somerset May 2 '12 at 7:57

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