# Examples of $G_\delta$ sets

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 ?

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Natural examples of Borel class higher than $G_\delta$ or $F_\sigma$ would be equally interesting. The experts were old masters (like Kazimierz Kuratowski) about a hundred years ago or so. –  Włodzimierz Holsztyński Oct 31 '14 at 5:21
@WłodzimierzHolsztyński That would be nicer. That, and a list of examples of functions of Baire class three. –  Andres Caicedo Oct 31 '14 at 6:01
Perhaps you (Andreas) or @user6129 may post the extension question? –  Włodzimierz Holsztyński Oct 31 '14 at 15:52

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.

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What a nice answer! –  mathahada Mar 5 '11 at 9:06
Also: in the Fréchet space $C^\infty(\mathbb{R})$, the set of all functions that are nowhere analytic is a dense $G_\delta$, a result by Morgenstern. –  Pietro Majer Oct 31 '14 at 9:30

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 $1/n$.)

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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.)

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That's interesting. Do you know a reference? –  arsmath Mar 4 '11 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 '11 at 13:26
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 '11 at 13:40

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

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This has a nice consequence: the set of all bounded functions $f'$ for which that set is dense is a Banach space under the uniform norm. –  Pietro Majer Oct 31 '14 at 11:40

Most mathematicians consider connected spaces that are not path-connected to be a pathology (and perhaps they are right). So, it may come as a surprise that most continua are actually not path-connected. The pseudo-arc is a non-degenerate (i.e. at least $2$ points), hereditarily indecomposable, chainable continuum. Chainable means that if $P$ is a pseudo-arc then, for every $\varepsilon>0$, $P$ is $\varepsilon$-close to an arc (i.e. a path). On the other hand being hereditarily indecomposable implies that the pseudo-arc is very much not path-connected: Each of its path-components is a singleton!

Bing has proved in Concerning hereditarily indecomposable continua, Pacific J. Math. V.1,N.1(1951), 43-51 http://projecteuclid.org/euclid.pjm/1102613150 that (the pseudo-arc is unique up to a homeomorphism, and) for any Euclidean space or Hilbert space $E$ of dimension greater than $1$ the collection of subcontinua of $E$ which are pseudo-arcs is a dense $G_\delta$ in the hyperspace of subcontinua of $E$.

W. Lewis later proved in Most maps of the pseudo-arc are homeomorphisms, Proc. Amer. Math. Soc. 91 (1984), no. 1, 147–154 http://www.ams.org/journals/proc/1984-091-01/S0002-9939-1984-0735582-4/ that: If $M(P)$ is the space of maps of the pseudo-arc into itself with the sup metric, then the subset $H(P)$ of maps of the pseudo-arc into itself which are homeomorphisms onto their images is a dense $G_\delta$ in $M(P)$.

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Here is an example by Roman Sikorski:

$$(H\setminus\{\mathbf 0\})\,^\mathbb N$$

in $\ H=_{top} H\,^\mathbb N,\$where $\ H\$ is the countably-dimensional Hilbert space (you can have the Hilbert cube instead). Of course any metric sopace would do, but this one is universal for $G_\delta$-sets in the Sikorski's sense. This is just one of the double transfinite sequence of universal sets for all Borel classes. Sikorski called Borel classes additive and productive, so that--for instance--$G_\delta$ would be productive of class 1.

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The set $\mathbb R\setminus\mathbb Q$ of all irrational numbers.

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That's what @user6129 said in their Question (the $G_\delta$ of continuity points of the second Dirichlet function). –  Włodzimierz Holsztyński Oct 31 '14 at 4:55

Let $E\subset\mathbb{R}^n$ be a Lebesgue measurable set: then there exists a $G_\delta$ set $G\supset E$ such that $m(G\setminus E)=0.$

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This is not an example, this is a theorem :-) –  Włodzimierz Holsztyński Oct 31 '14 at 4:52
Actually it is an example, coming from some result, like everyone else here (the $G_\delta$ is not explicit though). –  Pietro Majer Oct 31 '14 at 6:58
As long as we do not question the definition of the definition we are still in good shape. –  Włodzimierz Holsztyński Oct 31 '14 at 8:09
Any way I look at it, this answer says that an example of a $G_\delta$ set is: all of them. –  Emil Jeřábek Oct 31 '14 at 10:54
Say that given a concrete measurable set, there is a more or less concrete example of a $G_\delta$ cover of it. OK, not a big example, but the best ones were already gone. –  Pietro Majer Oct 31 '14 at 11:24

The set of $C^1$ diffeomorphisms of a compact manifold for which all periodic points are hyperbolic and their stable and unstable manifolds intersect transversely.

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Let $X$ be a metric space, $A \subset X$, and $f:A \to Y$ be continuous, where $Y$ is a complete metric space. The $f$ extends to $f^* : A^* \to Y$ for a $G_\delta$ subset $A^* \subset X$ s.t. $A \subset A^* \subset \overline{A}$. (Willard, Lemma 25.8)

Also, completely metrizable spaces are those which are $G_\delta$-sets in any metric space in which they are embedded. More exhaustively,

$X$ is completely metrizable $\iff$ $X$ is a $G_\delta$ in its completion $\iff$ $X$ is a $G_\delta$ in every metric embedding $\iff$ $X$ is a $G_\delta$ in $\beta X$ $\iff$ $X$ is a $G_\delta$ whenever embedded densely in a Tychonoff space. (Willard, Thm 24.13)

References are to the very best textbook I have ever seen on point set topology, Willard's General Topology, 1970.

(I was looking up the universal properties of completions for my number theory class, in order to avoid reinventing the house, when I came across these nice characterizations.)

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