Piecewise syndetic sets, topological dynamics and ultrafilters

In this post I explore the notion of piecewise syndeticity and its relation to topological dynamical systems and the Stone-Čech compactification. I restrict attention to the additive semigroup {({\mathbb N},+)} but most results presented are true in much bigger generality (and I tried to present the proofs in a way that does’t depend crucially on the fact that the semigroup is {({\mathbb N},+)}). These results and observations are collected from several sources, and are scattered in the literature so I thought it would be nice to collect them all in one place. Probably the union of Furstenberg’s book with the book by Hindman and Strauss contains most (if not all) of the things in here; these surveys of Bergelson are also a good source.

— 1. Introduction —

There are several notions of largeness for subsets of {{\mathbb N}}, each notion with its advantages and disadvantages. One such notion is that of syndetic set, i.e., a set {S\subset{\mathbb N}} with bounded gaps. More precisely, {S} is syndetic if there exists a finite set {F} such that any natural number can be written as {a-b} with {a\in S} and {b\in F}.

Example 1 The following are syndetic sets:

  • The set {{\mathbb N}} of all natural numbers,
  • Any co-finite subset {S\subset{\mathbb N}},
  • The set {2{\mathbb N}} of even numbers,
  • Any infinite progression of the form {a{\mathbb N}+b},
  • The set {\{n\in{\mathbb N}:d(n\alpha,{\mathbb N})<\epsilon\}} for any {\alpha,\epsilon>0}, where {d(x,{\mathbb N})} is the distance between the positive real number {x} and the lattice of natural numbers.
  • The set {\{n\in{\mathbb N}:d(f(n),{\mathbb N})<\epsilon\}} for any {\epsilon>0} and any {f\in{\mathbb R}[x]} with an irrational coefficient (other than the constant term).

One problem with syndeticity is that this notion is not partition regular, i.e. if {S} is a syndetic set and we decompose {S=A\cup B}, it may happen that neither {A} nor {B} are syndetic (take {S={\mathbb N}} and {A} and {B} consisting of increasingly long alternating sequences of intervals). One way to fix this is to consider its dual family and then its intersection family (spoiler alert: we end up with piecewise syndetic sets).

Definition 1 A set {T\subset{\mathbb N}} is called thick if it contains arbitrarily long intervals. More precisely, if for any finite set {F\subset{\mathbb N}} there exists {n\in{\mathbb N}} such that {F+n\subset T}.

It should be clear that it is possible for a thick set {T} to be decomposed into two sets {T=A\cup B} with neither {A} nor {B} being thick.

Example 2 The following are thick sets:

  • The set {{\mathbb N}} of all natural numbers,
  • Any co-finite subset {T\subset{\mathbb N}},
  • The set {\{n\in{\mathbb N}:\left|\hat\mu(n)\right|<\epsilon\}} where {\epsilon>0} and {\hat\mu} is the Fourier transform of a non-atomic measure {\mu} on {{\mathbb T}}.
  • The set {\{n:d(n^\alpha+\beta,{\mathbb N})<\epsilon\}} for any {\epsilon,\alpha,\beta>0} with {\alpha\notin{\mathbb N}}.

It is not hard to check that {T} is thick if and only if {T\cap S\neq\emptyset} for every syndetic set {S}. Given a family and its dual family it is natural to consider the family of intersections:

Definition 2 A set {A\subset{\mathbb N}} is called piecewise syndetic if there exist a syndetic set {S} and a thick set {T} such that {A=S\cap T}.

Example 3 Any syndetic set is piecewise syndetic and any thick set is piecewise syndetic. My earlier post listed several sets which are NOT syndetic; it turns out that the same proofs show that those sets are not piecewise syndetic. These examples (of sets which are not piecewise syndetic) include:

  • The set of prime numbers.
  • The set of square-free numbers (and more generally, given {k\in{\mathbb N}}, the set of numbers not divisible by any power {p^k} with {p} prime).
  • The set of numbers with at most {k} distinct prime divisors, for a fixed but arbitrary {k\in{\mathbb N}}.
  • Let {k\in{\mathbb N}}. The set of numbers {n\in{\mathbb N}} such that when {n} is decomposed into prime powers, there is no exponent equal to {k}.

It turns out that a family of intersections must be partition regular:

Proposition 3 Let {{\mathcal S}} be a family of subsets of {{\mathbb N}} such that if {S\in{\mathcal S}} and {R\subset{\mathbb N}} contains {S}, then also {R\in{\mathcal S}}. Let {{\mathcal T}:=\{T\subset{\mathbb N}:(\forall\; S\in{\mathcal S}) T\cap S\neq\emptyset\}} be the dual family of {{\mathcal S}}. Also, let {{\mathcal A}:=\{S\cap T:S\in{\mathcal S},\ T\in{\mathcal T}\}} be the intersection family.

Whenever a set {A\in{\mathcal A}} is decomposed into finitely many pieces {A=A_1\cup\cdots\cup A_r}, some piece {A_t} is also in {{\mathcal A}}.

Proof: It suffices to show the result for {r=2}; the general case follows by induction. Assume that {A=S\cap T}, where {S\in{\mathcal S}} and {T\in{\mathcal T}} is decomposed {A=B\cup C} with {B\cap C=\emptyset}. Let {\tilde S:=B\cup(S\setminus A)}. It is easy to check that {B=\tilde S\cap T}, so if {\tilde S\in{\mathcal S}} then {B\in{\mathcal A}} and the proof is finished. But if, on the contrary, {\tilde S\notin{\mathcal S}}, we have {\tilde T:={\mathbb N}\setminus\tilde S\in{\mathcal T}}. Finally observe that {C=\tilde T\cap S}, which shows that in this case {C\in{\mathcal A}}. \Box

In particular we obtain the following classical result.

Corollary 4 (Brown’s lemma) Let {A\subset{\mathbb N}} be a piecewise syndetic set and assume that {A=A_1\cup\cdots\cup A_r}. Then for some {t\in[r]:=\{1,\dots,r\}} we have that {A_t} is piecewise syndetic.

We remark that Proposition 3 does not appeal to any property of {{\mathbb N}}, and hence holds with any underlying set. In particular, Corollary 4 holds for any semigroup.

The following description of piecewise syndetic sets is not surprising but sometimes useful.

Lemma 5 A set {A\subset{\mathbb N}} is piecewise syndetic if and only if there exists a syndetic set {S\subset{\mathbb N}} such that for any finite subset {F\subset S} there exists a shift {m=m(F)\in{\mathbb N}} such that {F+m(F)\subset A}.


Let {F_1\subset{\mathbb N}} be a finite set so that {S-F_1={\mathbb N}}. Denote by {{\mathcal F}} the family of all finite non-empty subsets of {{\mathbb N}} and let {T:=\bigcup_{F\in{\mathcal F}}F+m((F+F_1)\cap S)} and let {\tilde S=A\cup({\mathbb N}\setminus T)}. It is clear that {A\supset\tilde S\cap T} and that {T} is thick, so it suffices to show that {\tilde S} is syndetic. We will show that {\tilde S-F_1={\mathbb N}} Let {n\in{\mathbb N}} be arbitrary. If {n\notin\tilde S}, then {n\in T}, hence there exists {F\in{\mathcal F}} such that {n\in F+m((F+F_1)\cap S)}. Let {\ell\in F} be such that {n-m((F+F_1)\cap S)+\ell\in S}. Therefore {n-m((F+F_1)\cap S)+\ell\in(F+F_1)\cap S}, hence {n+\ell\in((F+F_1)\cap S)+m((F+F_1)\cap S)} and thus {n+\ell\in A\subset\tilde S}.

We now prove the converse. Assume that {A} is a piecewise syndetic set and let {\tilde S\subset{\mathbb N}} be a syndetic set and {T\subset{\mathbb N}} be a thick set such that {A=\tilde S\cap T} and let {F\subset{\mathbb N}} be a finite set such that {\tilde S-F={\mathbb N}}. Let {F_1=F} and, for each {i>1} let {F_i=F_{i-1}+F}. For each {i\in{\mathbb N}}, choose {n_i\in{\mathbb N}} so that {n_i+F_i\subset T}. We are not going to construct sequences {(s_j)} in {{\mathbb N}} with each {s_j\in F_j} and a sequence of infinite sets {{\mathbb N}\supset I_1\supset I_2\supset\cdots} such that for any {j\in{\mathbb N}} and {i\in I_j} we have {n_i+s_j\in\tilde S}. Observe that as long as {j\geq i} we have have that {n_i+s_j\in T} and hence {n_i+s_j\in A}.

First find an infinite set {I_1\subset{\mathbb N}} and an element {\ell_1\in F} such that for every {i\in I_1} we have {n_i+\ell_1\in\tilde S}. Let {s_1:=\ell_1}. For each {j=2,3,\dots}, let {I_j\subset I_{j-1}} be an infinite set and {\ell_j\in F} be such that for each {i\in I_j} we have {n_i+\ell_1+\cdots+\ell_{j-1}+\ell_j\in\tilde S}, and make {s_j=s_{j-1}+\ell_j}. Finally take {S=\{s_1,s_2,\dots\}}.

It is clear that {S} is syndetic, since any two consecutive elements of {S} differ by at most {\max F}. Moreover, given any finite subset {G\subset S}, say {G\subset\{s_1,s_2,\dots,s_j\}} for some {j\in{\mathbb N}}. Then for any {i\in I_j} with {i>j} we have {n_i+s_t\in A} whenever {t\leq j}. This shows that {n_i+G\subset A}, finishing the proof. \Box

— 2. Relation with topological dynamics —

A topological dynamical system (for the purposes of this post) is a pair {(X,T)}, where {X} is a compact Hausdorff space and {T:X\rightarrow X} is a continuous function. For instance one can take {X=\{0,1\}^{\mathbb N}} with the product topology and {T:X\rightarrow X} being the left shift (more precisely if {x=(x_n)_{n\in{\mathbb N}}} is a point in {X}, then {Tx=(x_{n+1})_{n\in{\mathbb N}}}). Given a topological system {(X,T)} and a point {x\in X}, the orbit closure of {x} is the set {{\mathcal O}(x):=\overline{\{T^nx:n\in{\mathbb N}\}}}. Orbit closures are invariant under {T}, so for any point {x\in X} one can restrict {T} to {{\mathcal O}(x)} and obtain a new topological dynamical system which contains all the information about the point {x}.

An interesting feature of topological dynamical systems are the properties of the return times to a neighborhood. Given any point {x\in X} and a neighborhood {U\ni x}, what can be said of the set of return times {R(x,U):=\{n\in{\mathbb N}:T^nx\in U\}}? It turns out that properties of the dynamical systems are often reflected in the notion of largeness possessed by the sets {R(x,U)}.

Example 4

  • Let {X=[0,1]} and {T:X\rightarrow X} be the map {Tx=x^2}. There are exactly two fixed points: {0} and {1}. Given any point {x\in(0,1)} there is a neighborhood {U\ni x} such that the set {R(x,U)} is finite.
  • Let {X} be a compact group and let {\alpha\in X} be such that the subgroup of {X} generated by {\alpha} is dense in {X} (for instance, take {X={\mathbb R}/{\mathbb Z}} and {\alpha} to be irrational). Let {T:X\rightarrow X} be the rotation {T:x\mapsto x+\alpha}. Then for any {x\in X} and {U\ni x}, the set {R(x,U)} is a Bohr set and, in particular, syndetic.

A topological dynamical system {(X,T)} is called minimal if every point has a dense orbit, i.e., {{\mathcal O}(x)=X} for every {x\in X}. It is clear that a minimal non-empty closed invariant subset of {X} must be a minimal subsystem; therefore Zorn’s lemma guarantees that any topological dynamical system contains a minimal subsystem. It turns out that minimal systems are essentially characterized by the property that the sets of return times are syndetic.

Proposition 6 If a topological dynamical system {(X,T)} is minimal, then for any open set {U\subset X} and any point {x\in U} the set {R(x,U)} is syndetic.

Proof: Let {Y:=\bigcup_{\ell\in{\mathbb N}} T^{-\ell}U}, we claim that {Y=X}. Indeed, assume that {y\in X\setminus Y}, then {T^ny\notin U} for any {n\in{\mathbb N}} and hence the orbit of {y} can not be dense, contradicting the minimality. By compactness of {X}, there exists some {M\in{\mathbb N}} such that {X=\bigcup_{\ell=1}^MT^{-\ell}U}.

I claim that for any {m\in{\mathbb N}} there exists {\ell\in[M]} and {n\in R(x,U)} such that {m=n-\ell}, this will finish the proof. Indeed, let {m\in{\mathbb N}} and take {y=T^mx}. Let {\ell\in[M]} be such that {y\in T^{-\ell}U}. It follows that {T^{m+\ell}x\in U}, which implies that {n:=m+\ell\in R(x,U)}. \Box

There is a partial converse to the previous proposition, if one adds the assumption of transitivity:

Proposition 7 Let {(X,T)} be a topological dynamical system. If there exists one point {z\in X} with a dense orbit and every set of the form {R(x,U)} is syndetic, then the system is minimal.

Proof: We need to show that for any {x\in X} we have {{\mathcal O}(x)=X}. For this it suffices to show that {z\in{\mathcal O}(x)}, for this implies that {X={\mathcal O}(z)\subset{\mathcal O}(x)}. In order to show this we need to prove that given an arbitrary neighborhood {U} of {z}, the pre-orbit {\bigcup_{n\in{\mathbb N}}T^{-n}U=X}. Since compact Hausdorff spaces are normal, there exists a neighborhood {V} of {z} whose closure {\bar V\subset U}. The set {R(z,V)} is syndetic, so there exists a finite set {F\subset{\mathbb N}} such that for any {m\in{\mathbb N}} we can write {m=n-\ell} for some {n\in R(z,V)} and {\ell\in F}. Therefore {T^\ell(T^mz)=T^nz\in V} and hence {T^mz\in T^{-\ell}V}. Since {m\in{\mathbb N}} was arbitrary, we deduce that each point {T^mz} in the orbit of {z} belongs to {W:=\bigcup_{\ell\in F}T^{-\ell}V}, and hence {X={\mathcal O}(z)=\overline W}. Finally, observe that

\displaystyle X=\overline W=\overline{\bigcup_{\ell\in F}T^{-\ell}V}=\bigcup_{\ell\in F}\overline{T^{-\ell}V}\subset\bigcup_{\ell\in F}T^{-\ell}\overline V\subset\bigcup_{n\in{\mathbb N}}T^{-n}U


Now let {A\subset{\mathbb N}}. Its indicator function {1_A} is a point in the compact space {\{0,1\}^{\mathbb N}}. Several properties of the set {A} are encoded in the topological dynamical system {({\mathcal O}(1_A),T)}, where {T:\{0,1\}^{\mathbb N}\rightarrow\{0,1\}^{\mathbb N}} is the shift map. In particular, we have:

Proposition 8 A set {A\subset{\mathbb N}} is piecewise syndetic if and only if there exists a minimal subsystem of {({\mathcal O}(1_A),T)} which is not the fixed point {\{(0,0,\dots)\}}.

Proof: First assume that {A} is piecewise syndetic. Using Lemma 5 we can find a syndetic set {S\subset{\mathbb N}} such that for any finite piece {F\subset S} there exists a shift {m\in{\mathbb N}} such that {F+m\in A}. For each {n\in{\mathbb N}} let {m(n)\in{\mathbb N}} be such that {(S\cap[n])+m(n)\subset A}. This implies that {T^{m(n)}1_A(k)\geq1_S(k)} for each {k\in[n]}. Let {x\in X} be any limit point of the sequence {T^{m(n)}1_A}, the previous observation shows that {x\geq1_S} and hence {x} is the indicator function of a syndetic set. Let {Y} be any minimal system contained in the orbit closure of {x}, observe that {Y\subset{\mathcal O}(x)\subset{\mathcal O}(1_A)}. To finish the proof of this implication we need to prove that the point {(0,0,\dots)} can not belong to {{\mathcal O}(x)}.

Indeed, let {R:=\{n\in{\mathbb N}:x(n)=1\}} and take {F\subset{\mathbb N}} such that {{\mathbb N}=R-F}. Let {U:=\{y\in\{0,1\}^{\mathbb N}:(\forall\; n\in F)y(n)=0\}}; this is a neighborhood of {(0,0,\dots)}. Assume, for the sake of a contradiction, that {T^nx\in U} for some {n\in{\mathbb N}}. Then {x(n+m)=0} for every {m\in F}, contradicting the fact that {n=r-m} for some {r\in R} and {m\in F}. This contradiction ends the proof of the first implication.

Next we prove the converse: assume that such minimal subsystem {(X,T)} exists and let {U=\{y\in X:y(1)=1\}}. Observe that {U} is a non-empty open subset of {X}. Let {x\in U} be arbitrary and let {S=\{n\in{\mathbb N}:T^nx\in U\}=R(x,U)}. In view of Proposition 6, {S} is syndetic. We will show that any finite subset of {S} can be shifted into {A}; in view of Lemma 5 this will finish the proof.

Let {F\subset S} be an arbitrary finite set and let {V:=\{y\in X:(\forall\;n\in F)y(n+1)=1\}}. Observe that {V} is an open neighborhood of {x}. Since {x\in{\mathcal O}(1_A)}, there exists some {m\in{\mathbb N}} such that {T^m1_A\in V}. Unwrapping the definitions, this means that {F+m\subset A} as desired. \Box

Proposition 8 provides a way to relate a combinatorial property of a set of integers and a property of a dynamical system naturally constructed from {A}. This relation is made more precise with the notion of sets of recurrence.

Definition 9 A set {R\subset{\mathbb N}} is called a set of topological recurrence if for any minimal system {(X,T)} and any non-empty open set {U\subset X} there exists {n\in R} such that {U\cap T^{-n}U\neq\emptyset}.

The notion of sets of recurrence in dynamical systems is closely related to Ramsey theory on {{\mathbb N}}. I have mentioned sets of recurrence before in this blog, and in particular mentioned their relation with sets of chromatic recurrence. We can now add another equivalent characterization.

Theorem 10 Let {R\subset{\mathbb N}}. The following are equivalent:

  1. {R} is a set of topological recurrence.
  2. {R} is a set of chromatic recurrence (i.e., whenever {{\mathbb N}=C_1\cup\cdots\cup C_r} there exists {t\in[r]} and {n\in R} such that {C_t\cap(C_t-n)\neq\emptyset}).
  3. For any syndetic set {S\subset{\mathbb N}} there exists {n\in R} such that {S\cap(S-n)\neq\emptyset}.
  4. For any piecewise syndetic set {A\subset{\mathbb N}} there exists {n\in R} such that {A\cap(A-n)\neq\emptyset}.

I will mention in passing that it is a famous question of Katzenelson whether an equivalent formulation is obtained when one replaces syndetic (or piecewise syndetic) sets by Bohr sets.

Proof: A proof of the equivalence between parts (1) and (2) was presented in my previous post. In view of Corollary 4 we have that (4) implies (2). Since finitely many shifts of a syndetic set form a finite partition of {{\mathbb N}}, we have that (2) implies (3), therefore it suffices to show that (1) implies (4).

Assume that (1) holds for a set {R\subset{\mathbb N}}, let {A\subset{\mathbb N}} be a piecewise syndetic and, invoking Proposition 8, let {X\in{\mathcal O}(1_A)} be a non-zero minimal subsystem. Let {U:=\{x\in X:x(1)=1\}}, observe that {U} is non-empty and let {n\in R} be such that {V:=U\cap T^{-n}U\neq\emptyset}. Since {V} is a non-empty open subset of {{\mathcal O}(1_A)}, there exists {m\in{\mathbb N}} such that {T^m1_A\in V}, and in particular both {m+1} and {m+n+1} belong to {A}, showing that {m+1\in A\cap(A-n)}. \Box

— 3. Relation with minimal ultrafilters —

I have written before in this blog about a way to relate ultrafilters and piecewise syndetic sets; namely presenting a lovely proof of Beiglböck of a theorem of Jin (stating that whenever {A,B\subset{\mathbb N}} have positive Banach upper density, their sum {A+B} is piecewise syndetic). This section will be (almost) unrelated to that proof!

There is an interesting relation between piecewise syndetic sets and certain special points in the Stone-Čech compactification {\beta{\mathbb N}} of {{\mathbb N}}. In this previous post I presented the construction of the {\beta{\mathbb N}} as the set of all ultrafilters over {{\mathbb N}}. An ultrafilter is a non-empty family {p} of subsets of {{\mathbb N}} satisfying the following (somewhat redundant) properties:

  1. {{\mathbb N}\in p}, but {\emptyset\notin p}.
  2. If {A\in p} and {B\supset A}, then {B\in p}.
  3. If {A,B\in p} then also {A\cap B\in p}.
  4. We have {A\in p} if and only if {({\mathbb N}\setminus A)\notin p}.
  5. If {A\in p} and we decompose {A=A_1\cup\cdots\cup A_r} into finitely many disjoint sets, then for exactly one {t\in[r]} we have {A_t\in p}.

The topology on the set {\beta{\mathbb N}} of all ultrafilters over {{\mathbb N}} is generated by the clopen sets {\bar A:=\{p\in\beta{\mathbb N}:A\in p\}} whenever {A\subset{\mathbb N}} is non-empty. With this topology, {\beta{\mathbb N}} becomes a compact Hausdorff (but not metrizable) space. There is a natural action from {({\mathbb N},+)} on {\beta{\mathbb N}} given by {T:p\mapsto p+1:=\{A\subset{\mathbb N}:A-1\in p\}}. We just created a topological dynamical system {(\beta{\mathbb N},T)}, as in the previous section this system has minimal subsystems. The following proposition characterizes piecewise syndetic sets in terms of minimal subsystems of {(\beta{\mathbb N},T)}.

Proposition 11 A set {A\subset{\mathbb N}} is piecewise syndetic if and only if there exists an ultrafilter {p\in\beta{\mathbb N}} which belongs to a minimal subsystem of {(\beta{\mathbb N},T)} and such that {A\in p}.

Proof: First assume that {p} belongs to a minimal subsystem {X} of {(\beta{\mathbb N},T)} and let {A\in p}. This means that {p\in\bar A}, and hence Proposition 6 tells us that {R(p,\bar A)} is syndetic. I claim that any finite subset of {R(p,\bar A)} can be shifted into a subset of {A}, which in view of Lemma 5 will finish the proof of this implication.

Indeed, let {F\subset R(p,\bar A)}. Unwrapping the definition we obtain {A-n\in p} for each {n\in F}, and hence the finite intersection {B=\bigcap_{n\in F}(A-n)\in p} is non-empty. Take any {m\in B}, it follows that {m+F\subset A} as claimed.

Next we prove the converse direction. Assume {A} is a piecewise syndetic set and let {S} and {H} be, respectively, a syndetic and a thick set such that {A=S\cap H}. For each finite set {F\subset{\mathbb N}}, let {H_F:=\{m\in{\mathbb N}:m+F\subset H\}}. Since {H} is thick, {H_F\neq\emptyset}. Moreover, given any finitely many finite sets {F_1,\dots F_r} in {{\mathbb N}}, the intersection {H_{F_1}\cap\cdots H_{F_r}=H_{F_1\cup\cdots\cup F_r}\neq\emptyset}. Therefore the compact subsets {\{\overline{H_F}:F\subset{\mathbb N};|F|<\infty\}} have the finite intersection property and hence have a non-empty intersection {B:=\bigcap_{F\subset{\mathbb N},|F|<\infty}H_F\neq\emptyset}. It is not hard to check that for any {p\in B} and {n\in{\mathbb N}} we have {p+n\in B}; therefore {(B,T)} is a subsystem of {(\beta{\mathbb N},T)}. Let {X} be a minimal subsystem of {B}.

Let {F\subset{\mathbb N}} be such that {S-F={\mathbb N}}. Therefore for some {n\in F} we have {\overline{(S-n)}\cap X\neq\emptyset}. Let {q} be in that intersection and let {p=q+n}. It follows that {p\in X} and {S\in p}. But since {p\in X} we also have that {H\in p}, therefore the intersection {A=S\cap H\in p} and this finishes the proof. \Box

Observe that Corollary 4 also follows directly from this proposition.

Proposition 11 can be used to show that given any piecewise syndetic set {A\subset{\mathbb N}} there exists a shift {A-m} (for some {m\in{\mathbb N}}) which is central (the definition of central set is explored, for instance, in this survey by Bergelson) and in my previous post (it’s Definition 4 there).

This entry was posted in Classic results, Combinatorics, Tool, Topological Dynamics and tagged , , . Bookmark the permalink.

3 Responses to Piecewise syndetic sets, topological dynamics and ultrafilters

  1. Pingback: An arithmetic van der Corput trick and the polynomial van der Waerden theorem | I Can't Believe It's Not Random!

  2. Pingback: Single and multiple recurrence along non-polynomial sequences | I Can't Believe It's Not Random!

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