The time has come to reveal what the experiment in the previous post was about. As with many experiments (the most famous probably being that of Stanley Milgram about obedience to authority), its real purpose was not its ostensive purpose.
Over the last three years, I have been collaborating with Mohan Ganesalingam, a computer scientist, linguist and mathematician (and amazingly good at all three) on a project to write programs that can solve mathematical problems. We have recently produced our first program. It is rather rudimentary: the only problems it can solve are ones that mathematicians would describe as very routine and not requiring any ideas, and even within that class of problems there are considerable restrictions on what it can do; we plan to write more sophisticated programs in the future. However, each of the problems in the previous post belongs to the class of problems that it can solve, and for each problem one write-up was by an undergraduate mathematician, one by a mathematics PhD student and one by our program. (To be clear, the program was given the problems and produced the proofs and the write-ups with no further help. I will have more to say about how it works in future posts.) We wanted to see whether anybody would suspect that not all the write-ups were human-generated. Nobody gave the slightest hint that they did.
Of course, there is a world of difference between not noticing a difference that you have not been told to look out for, and being unable to detect that difference at all. Our aim was merely to be able to back up a claim that our program produces passable human-style output, so we did not want to subject that output to full Turing-test-style scrutiny, but you may, if you were kind enough to participate in the experiment, feel slightly cheated. Indeed, in a certain sense you were cheated — that was the whole point. It seems only fair to give you the chance to judge the write-ups again now that you know how they were produced. For each problem I have created a poll, and each poll has seven possible answers. These are:
The computer-generated output is definitely (a).
I think the computer-generated output is (a) but am not certain.
The computer-generated output is definitely (b).
I think the computer-generated output is (b) but am not certain.
The computer-generated output is definitely (c).
I think the computer-generated output is (c) but am not certain.
I have no idea which write-up was computer generated.
I would also be interested in comments about how you came to your judgments. All comments on both experiments and all votes in the polls will be kept private until I decide that it is time to finish the second experiment. A small remark is that I transcribed by hand all the write-ups into a form suitable for WordPress, so the existence of a typo in a write-up is not a trivial proof that it was by a human.
If you did not participate in the first experiment but nevertheless want to try this one, that’s fine. [Update: I have now closed the polls. Very soon Mohan and I will post the results and a discussion of them. Further update: The results now appear below. They appear displayed in possibly a more convenient way in this post, which also contains a discussion of how the program works.]
Problem 1. Let and be open sets in a metric space. Then is open.
1(a) We want to show that for all in , there exists such that the open ball is contained in .
Let . is in , so is in and is in . Since and are open, there exist such that and . Let . Then is contained in , so it’s contained in . Similarly, is contained in , so it’s contained in . So . So is open.
1(b) For arbitrary , let . Consider an arbitrary . As are open there are such that and . Take . Then and . So . We’ve proved that for any there is an open ball ( in this case) that contains and is inside . So is open.
1(c) Let be an element of . Then and . Therefore, since is open, there exists such that whenever and since is open, there exists such that whenever . We would like to find s.t. whenever . But if and only if and . We know that whenever and that whenever . Assume now that . Then if and if . We may therefore take and we are done.
Problem 2. Let and be metric spaces, let be continuous, and let be an open subset of . Then is an open subset of .
2(a) Let be an element of . Then . Therefore, since is open, there exists such that whenever . We would like to find s.t. whenever . But if and only if . We know that whenever . Since is continuous, there exists such that whenever . Therefore, setting , we are done.
2(b) Let . We seek such that the open ball is contained in .
, so . is open, so we know that for some , . Since is continuous, there exists such that for all , ; i.e., . So if ; i.e., . So is open.
2(c) Take any . We have . As is open, there is an open ball in . Because is continuous, there is some such that for any , belongs to . Hence, for such , . So . So . We’ve proved that every point in has an open ball neighbourhood. So is open.
Problem 3. Let be a complete metric space and let be a closed subset of . Then is complete.
3(a) Consider an arbitrary Cauchy sequence in . As is complete, has a limit in . Suppose . Because is closed, belongs to . We’ve proved that every Cauchy sequence in has a limit point in . So is complete.
3(b) Let be a Cauchy sequence in . Then, since is complete, we have that converges. That is, there exists such that . Since is closed in , is a sequence in and , we have that . Thus converges in and we are done.
3(c) Let be a Cauchy sequence in . We want to show that tends to a limit in .
Since is a subset of , is a Cauchy sequence in . Since is complete, , for some . Since is a closed subset of , it must contain all its limit points, so . So in . So is complete.
Problem 4. Let and be metric spaces and let and be continuous. Then the composition is continuous.
4(a) Let , and let . We need to show that there exists such that for all , .
is continuous, so there exists such that for all , . is continuous, so there exists such that for all , . But then , as desired. So is continuous.
4(b) Take an arbitrary . Let and . Using continuity of , for any , there is some such that if (for ), then . As is continuous, there is some such that if (for ), then . So for any we’ve found such that if , then and therefore . Hence is continuous.
4(c) Take and . We would like to find s.t. whenever . Since is continuous, there exists such that whenever . Since is continuous, there exists such that whenever . Therefore, setting , we are done.
Problem 5. Let and be sets, let be an injection and let and be subsets of . Then .
5(a) Take . So there is some and such that . As is injective, and are equal. So . So .
5(b) Suppose . Then, for some , and . So . Since is injective, , so , so . So .
5(c) Let be an element of . Then and . That is, there exists such that and there exists such that . Since is an injection, and , we have that . We would like to find s.t. . But if and only if and . Therefore, setting , we are done.