Is there a name for sets for which it is easier to test membership than to find members---and vice versa? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:09:34Z http://mathoverflow.net/feeds/question/42219 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42219/is-there-a-name-for-sets-for-which-it-is-easier-to-test-membership-than-to-find-m Is there a name for sets for which it is easier to test membership than to find members---and vice versa? Dick Palais 2010-10-14T21:47:17Z 2011-01-07T15:23:05Z <p>This is a question my son Bob asked me. For some sets it is relatively easy to test for membership but a lot more difficult to find members, and for others the reverse is true. Here is an elementary example to get the idea across. An $m \times n$ real matrix $M$ defines a linear map $x \mapsto M x = y$, from ${\mathbb R}^n$ to ${\mathbb R}^m$. It is easy to test if $x$ is in the kernel; just compute $M x$ and see if it is zero, but to find an $x$ in the kernel you must solve $M x = 0$ which is more computationally intensive. Conversely it is easy to find an element in the range; just choose any $x$ and compute $M x$; but to test if $y$ is in the range you must solve $M x = y$. Does anyone know if there is a standard name for this distinction or for sets of these two types?</p> http://mathoverflow.net/questions/42219/is-there-a-name-for-sets-for-which-it-is-easier-to-test-membership-than-to-find-m/42232#42232 Answer by Joel David Hamkins for Is there a name for sets for which it is easier to test membership than to find members---and vice versa? Joel David Hamkins 2010-10-14T23:34:41Z 2010-10-15T00:49:03Z <p>This phenomenon occurs both positively and negatively in many parts of logic, but to my knowledge, there is no particular adjective that is always used in such situations.</p> <ul> <li>In classical <em>computability theory</em>, the first phenomenon does not occur. If one can computably test membership in a set, in the usual Turing sense, then one can computably generate an instance, simply because one can computably enumerate all objects in the domain of discourse, and systematically test them. This is related to the classical fact that if the graph of a function is decidable, then the function is computable.</li> </ul> <p>Thus, to my mind, the phenomenon is intimately wrapped up with the ability to effectively enumerate, in the relevant sense, the objects in the domain of discourse.</p> <ul> <li><p>The converse situation, however, does occur in computability theory, and is a central phenomenon. Namely, there are sets of natural numbers whose members can be systematically generated---so the set is computably enumerable---but whose membership test is not computable. These are exactly the sets that are c.e. but not computable. Examples would include the halting problem (the set of programs $e$ halting on trivial input) and many other examples. It is easy to generate many halting programs---one can systematically enumerate them---but impossible to test in general if a given program halts. There is an intensively-studied hierarchy of Turing degrees instantiated by c.e. sets that are not decidable.</p></li> <li><p>In <em>complexity theory</em>, there is a sense in which there are negative examples. One can imagine a polynomial-time decidable set $A$, all of whose members are very large, and hence difficult to produce. To make the problem precise, however, one should really have a sequence $A_n$ of sets such that membership $x\in A_n$ is polynomial time decidable in $(x,n)$---that is, uniformly in $n$---but such that there is no polynomial time computable function $f$ such that $f(n)\in A_n$. Such an example is provided simply by the sets $A_n$ consisting of all numbers at least $2^n$. Given a pair $(x,n)$, it is polynomial-time decidable in $(x,n)$ whether $x\geq 2^n$, but there is no polynomial function exceeding $n\mapsto 2^n$.</p></li> </ul> <p>Similar examples would be provided by any sequence of sets $A_n$, all of whose members were very large in comparison with $n$, but such that the membership problem $x\in A_n$ is easily decided.</p> <ul> <li><p>In various sorts of higher computability theory, there are additional negative instances. For example, with the theory of infinite time Turing machines, there are infinite time decidable sets of reals with no computable members. Indeed, the Lost Melody Theorem asserts precisely that there are infinite time decidable singletons $\{c\}$, such that the real $c$ is not writable by any infinite time Turing machine. That is, there are reals $c$, such that it is decidable by infinite time Turing machines whether a given real $x$ is $c$ or not, by no such machine can produce $c$ on its own. This seems to be the essence of your phenomenon. (The lost melody'' terminology arises from the situation, where a person is able to recognize a given melody when someone else sings it, but is unable to sing it on their own.)</p></li> <li><p>In descriptive set theory, one would look at whether a set of reals at a given level in the descriptive set-theoretic hierarchy has members at that same level. This is false in general, although there are special circumstances (some involving large cardinal hypotheses) in which instances of it are true. One way to look at it is as a Choice principle: given a subset $A$ of the plane $\mathbb{R}\times\mathbb{R}$, can one find a function $f$ of the same complexity with $\text{dom}(f)=\text{dom}(A)$ such that $(x,f(x))\in A$ for all $x\in\text{dom}(A)$? This problem is also known as the <em>uniformization</em> problem.</p></li> <li><p>In a more general set theoretic setting, it is natural to consider the situation of ordinal-definable sets. Does every non-empty ordinal definable set contain an ordinal-definable member? This turns out to be equivalent to the assertion known as $V=HOD$, which is independent of ZFC, as explained in the edited version of <a href="http://mathoverflow.net/questions/10413/definable-collections-without-definable-members-in-zf/10415#10415" rel="nofollow">this MO answer</a>. The reason is that the set of non-ordinal-definable sets of minimal rank is ordinal definable.</p></li> </ul> http://mathoverflow.net/questions/42219/is-there-a-name-for-sets-for-which-it-is-easier-to-test-membership-than-to-find-m/51402#51402 Answer by Timothy Chow for Is there a name for sets for which it is easier to test membership than to find members---and vice versa? Timothy Chow 2011-01-07T15:23:05Z 2011-01-07T15:23:05Z <p><a href="http://cstheory.stackexchange.com/questions/1848/easy-decision-problem-hard-search-problem" rel="nofollow">A similar question</a> was asked at cstheory.stackexchange; there are several examples from computational complexity given there. However, I agree that there seems to be no standard name for such things.</p>