VizualBusiness: A second experiment concerning mathematical writing.

Wednesday, April 3, 2013

A second experiment concerning mathematical writing.

A second experiment concerning mathematical writing.:
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.

Problem 1. Let $A$ and $B$ be open sets in a metric space. Then $A\cap B$ is open.
1(a) We want to show that for all $x$ in $A \cap B$ , there exists $r$ such that the open ball $B_x(r)$ is contained in $A \cap B$ .
Let $x \in A \cap B$ . $x$ is in $A \cap B$ , so $x$ is in $A$ and $x$ is in $B$ . Since $A$ and $B$ are open, there exist $a, b$ such that $B_a(x) \subset A$ and $B_b(x) \subset B$ . Let $r = \min\{a, b\}$ . Then $B_r(x)$ is contained in $B_a(x)$ , so it’s contained in $A$ . Similarly, $B_r(x)$ is contained in $B_b(x)$ , so it’s contained in $B$ . So $B_r(x) \subset A \cap B$ . So $A \cap B$ is open.
1(b) For arbitrary $x\in X$ , let $B_r(x):=\{y\in X|d(y,x)<r\}$ . Consider an arbitrary $x\in A\cap B$ . As $A,B$ are open there are $r,r'>0$ such that $B_r(x)\subset A$ and $B_{r'}(x)\subset B$ . Take $R=\min(r,r')$ . Then $B_R(x)\subset B_r(x)\subset A$ and $B_R(x)\subset B_{r'}(x)\subset B$ . So $B_R(x)\subset A\cap B$ . We’ve proved that for any $x\in A\cap B$ there is an open ball ( $B_R(x)$ in this case) that contains $x$ and is inside $A\cap B$ . So $A\cap B$ is open.
1(c) Let $x$ be an element of $A\cap B$ . Then $x\in A$ and $x\in B$ . Therefore, since $A$ is open, there exists $\eta>0$ such that $u\in A$ whenever $d(x,u)<\eta$ and since $B$ is open, there exists $\theta>0$ such that $v\in B$ whenever $d(x,v)<\theta$ . We would like to find $\delta>0$ s.t. $y\in A\cap B$ whenever $d(x,y)<\delta$ . But $y\in A\cap B$ if and only if $y\in A$ and $y\in B$ . We know that $y\in A$ whenever $d(x,y)<\eta$ and that $y\in B$ whenever $d(x,y)<\theta$ . Assume now that $d(x,y)<\delta$ . Then $d(x,y)<\eta$ if $\delta\leq\eta$ and $d(x,y)<\theta$ if $\delta\leq\theta$ . We may therefore take $\delta=\min\{\eta,\theta\}$ and we are done.

Take Our Poll
Problem 2. Let $X$ and $Y$ be metric spaces, let $f:X->Y$ be continuous, and let $U$ be an open subset of $Y$ . Then $f^{-1}(U)$ is an open subset of $X$ .
2(a) Let $x$ be an element of $f^{-1}(U)$ . Then $f(x)\in U$ . Therefore, since $U$ is open, there exists $\eta>0$ such that $u\in U$ whenever $d(f(x),u)<\eta$ . We would like to find $\delta>0$ s.t. $y\in f^{-1}(U)$ whenever $d(x,y)<\delta$ . But $y\in f^{-1}(U)$ if and only if $f(y)\in U$ . We know that $f(y)\in U$ whenever $d(f(x),f(y))<\eta$ . Since $f$ is continuous, there exists $\theta>0$ such that $d(f(x),f(y))<\eta$ whenever $d(x,y)<\theta$ . Therefore, setting $\delta=\theta$ , we are done.
2(b) Let $x \in f^{-1}(U)$ . We seek $r > 0$ such that the open ball $B_x(r)$ is contained in $f^{-1}(U)$ .
$x \in f^{-1}(U)$ , so $f(x) \in U$ . $U$ is open, so we know that for some $\epsilon > 0$ , $B_{\epsilon}(f(x)) \subset U$ . Since $f$ is continuous, there exists $\delta > 0$ such that for all $z \in X$ , $d(x, z) < \delta\implies d(f(x), f(z)) < \epsilon$ ; i.e., $f(z) \in B_{\epsilon}(f(x)) \subset U$ . So $z \in f^{-1}(U)$ if $d(x, z) < \delta$ ; i.e., $B_{\delta}(x) \subset f^{-1}(U)$ . So $f^{-1}(U)$ is open.
2(c) Take any $x\in f^{-1}(U)$ . We have $f(x)\in U$ . As $U$ is open, there is an open ball $B_\epsilon(f(x))$ in $U$ . Because $f$ is continuous, there is some $\delta>0$ such that for any $y\in B_\delta(x)$ , $f(y)$ belongs to $B_\epsilon(f(x))$ . Hence, for such $y$ , $f(y)\in U$ . So $y\in f^{-1}(U)$ . So $B_\delta(x)\subset f^{-1}(U)$ . We’ve proved that every point in $f^{-1}(U)$ has an open ball neighbourhood. So $f^{-1}(U)$ is open.

Take Our Poll
Problem 3. Let $X$ be a complete metric space and let $A$ be a closed subset of $X$ . Then $A$ is complete.
3(a) Consider an arbitrary Cauchy sequence $(x_n)_{\{n\in\mathbb{N}\}}$ in $A$ . As $X$ is complete, $(x_n)$ has a limit in $X$ . Suppose $\lim_{n\to\infty}x_n=x$ . Because $A$ is closed, $x$ belongs to $A$ . We’ve proved that every Cauchy sequence in $A$ has a limit point in $A$ . So $A$ is complete.
3(b) Let $(a_n)$ be a Cauchy sequence in $A$ . Then, since $X$ is complete, we have that $(a_n)$ converges. That is, there exists $a$ such that $a_n\to a$ . Since $A$ is closed in $X$ , $(a_n)$ is a sequence in $A$ and $a_n\to a$ , we have that $a\in A$ . Thus $(a_n)$ converges in $A$ and we are done.
3(c) Let $(a_n)$ be a Cauchy sequence in $A$ . We want to show that $(a_n)$ tends to a limit in $A$ .
Since $A$ is a subset of $X$ , $(a_n)$ is a Cauchy sequence in $X$ . Since $X$ is complete, $a_n \to a$ , for some $a \in X$ . Since $A$ is a closed subset of $X$ , it must contain all its limit points, so $a \in A$ . So $a_n \to a$ in $A$ . So $A$ is complete.

Take Our Poll
Problem 4. Let $X, Y$ and $Z$ be metric spaces and let $f:X\to Y$ and $g:Y\to Z$ be continuous. Then the composition $g\circ f$ is continuous.
4(a) Let $x \in X$ , and let $\epsilon > 0$ . We need to show that there exists $\delta>0$ such that for all $p\in X$ , $d(x, p) < \delta\implies d(g\circ f(x), g\circ f(p)) < \epsilon$ .
$g$ is continuous, so there exists $\gamma > 0$ such that for all $y \in Y$ , $d(y, f(x)) < \gamma\implies d(g(y), g\circ f(x)) < \epsilon$ . $f$ is continuous, so there exists $\delta > 0$ such that for all $p\in X$ , $d(x, p) < \delta\implies d(f(x), f(p)) < \gamma$ . But then $d(g\circ f(p), g\circ f(x)) <\epsilon$ , as desired. So $g\circ f$ is continuous.
4(b) Take an arbitrary $x\in X$ . Let $y=f(x)$ and $z=g(y)$ . Using continuity of $g$ , for any $\epsilon>0$ , there is some $\epsilon'$ such that if $d(y',y)<\epsilon'$ (for $y'\in Y$ ), then $d(f(y'),z)<\epsilon$ . As $f$ is continuous, there is some $\delta>0$ such that if $d(x,x')<\delta$ (for $x'\in X$ ), then $d(f(x'),y)<\epsilon'$ . So for any $\epsilon>0$ we’ve found $\delta>0$ such that if $d(x,x')<\delta$ , then $d(f(x),f(x'))<\epsilon'$ and therefore $d(g\circ f(x),g\circ f(x'))<\epsilon$ . Hence $g\circ f$ is continuous.
4(c) Take $x$ and $\epsilon>0$ . We would like to find $\delta>0$ s.t. $d(g(f(x)),g(f(y)))<\epsilon$ whenever $d(x,y)<\delta$ . Since $g$ is continuous, there exists $\eta>0$ such that $d(g(f(x)),g(f(y)))<\epsilon$ whenever $d(f(x),f(y))<\eta$ . Since $f$ is continuous, there exists $\theta>0$ such that $d(f(x),f(y))<\eta$ whenever $d(x,y)<\theta$ . Therefore, setting $\delta=\theta$ , we are done.

Take Our Poll
Problem 5. Let $X$ and $Y$ be sets, let $f:X\to Y$ be an injection and let $A$ and $B$ be subsets of $X$ . Then $f(A)\cap f(B) \subset f(A\cap B)$ .
5(a) Take $x\in f(A)\cap f(B)$ . So there is some $y\in A$ and $z\in B$ such that $f(y)=f(z)=x$ . As $f$ is injective, $y$ and $z$ are equal. So $y\in A\cap B$ . So $x=f(y)\in f(A\cap B)$ .
5(b) Suppose $y \in f(A) \cap f(B)$ . Then, for some $a \in A, b \in B$ , $y = f(a)$ and $y = f(b)$ . So $f(a) = f(b)$ . Since $f$ is injective, $a = b$ , so $a \in A \cap B$ , so $y \in f(A \cap B)$ . So $f(A) \cap f(B) \subset f(A \cap B)$ .
5(c) Let $x$ be an element of $f(A)\cap f(B)$ . Then $x\in f(A)$ and $x\in f(B)$ . That is, there exists $y\in A$ such that $f(y)=x$ and there exists $z\in B$ such that $f(z)=x$ . Since $f$ is an injection, $f(y)=x$ and $f(z)=x$ , we have that $y=z$ . We would like to find $u\in A\cap B$ s.t. $f(u)=x$ . But $u\in A\cap B$ if and only if $u\in A$ and $u\in B$ . Therefore, setting $u=y$ , we are done.

Take Our Poll