MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
4

3

Recall that a subset A of a metric space X is a $G_\delta$ subset if it can be written as a countable intersection of open sets. This notion is related to the Baire category theorem. Here are three examples of interesting sets that are $G_\delta$ sets.

  • The set of continuity points of a function $f:X\rightarrow R$.
  • The set of positively recurrent points of a continuous transformation $T: X \rightarrow X$. Recall that a point is recurrent if there exists some sequence $n_i\rightarrow \infty$ such that $T^{n_i}x \rightarrow x$.
  • The set of transitive points of a continuous transformation $T: X \rightarrow X$. Recall that a point is transitive if its orbit $\{T^n(x)\}_{n\in Z}$ is dense in $X$.

Question : Can you provide more examples of interesting $G_\delta$ sets ?

flag

4 Answers

5

The space of nowhere differentiable continuous functions $f: [0, 1] \to \mathbb{R}$, as a subspace of the space of all continuous functions under the sup-norm topology.

link|flag
What a nice answer! – mathahada Mar 5 2011 at 9:06
4

The set of oracles A (as elements of the product space $2^\omega$) such that $\mathrm P^A\ne\mathrm{NP}^A$. (This also works for nonequality of other oracle complexity classes.)

link|flag
That's interesting. Do you know a reference? – arsmath Mar 4 2011 at 13:10
Not really, but it follows from the simple observation that it's a $\Pi^0_2(A)$-property: for every clocked poly-time oracle Turing machine $M$, there exists an input $x$ such that $M^A$ fails to correctly compute $x\stackrel?\in\mathrm{SAT}^A$. – Emil Jeřábek Mar 4 2011 at 13:26
1 
Incidentally, the set is not $\Sigma^0_2(A)$, and this has a neat topological proof: since the set is invariant under a finite change of the oracle, and it's a nonempty proper subset of $2^\omega$, both the set and its complement have empty interior. Thus, if the set were $\Sigma^0_2(A)$, i.e., $F_\sigma$, then both the set and its complement would be meager, contradicting the Baire category theorem. – Emil Jeřábek Mar 4 2011 at 13:40
2

In the space of all subsets of $\mathbb{N}$ (identified via characteristic functions with $2^{\mathbb{N}}$ and topologized as the product of copies of the discrete 2-point space), the set of infinite subsets of $\mathbb{N}$ is a $G_\delta$ set. So is the set of those $A\subseteq\mathbb{N}$ for which $\sum_{n\in A}1/n$ diverges. (The same goes with other sequences in place of $1/n$.)

link|flag
1

Let $f\colon\mathbb{R}\to\mathbb{R}$ be differentiable. The set of points where $f'$ vanishes is a $G_\delta$.

link|flag

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.