# Let P be a polygon

I've encountered the following misunderstanding. I pose a question (to undergraduates in the U.S.), for example:

Let $$P$$ be a polygon of $$n$$ vertices. Is it true that every triangulation of $$P$$ has the same number of triangles?

This question depends on what constitutes a "triangulation," but assume the student know that. The answer is Yes: every triangulation of $$P$$ consists of $$n-2$$ triangles.

Here is the problem I encounter. The students apparently don't understand that "Let $$P$$ be a polygon" means, let your mind run over all possible polygons, so $$P$$ is an "arbitrary" polygon, in that it can be anything that fits the definition of a polygon (which the students also know). They wonder, well, maybe $$P$$ is a convex polygon, and should I answer specific to convex polygons?

This example doesn't quite illustrate the problem because the answer is always Yes. But when the answer is sometimes Yes, sometimes No, they seem to get confused over the quantifier. I think it may come down to the meaning of the phrase: "Let $$A$$ be a $$B$$." Let $$p$$ be a point in the plane $$\mathbb{R}^2$$—meaning any point in the plane, an "arbitrary" point in the plane. Let $$P$$ be a polygon, meaning any polygon. Let $$T$$ be a triangulation of a set of $$n$$ points. Does every triangulation of $$n$$ points have the same number of triangles? (No, not always.)

Q. Have you encountered this confusion in your teaching? If so, how do you circumvent it?

• Reminds me of this question I asked: matheducators.stackexchange.com/questions/13594/… Perhaps we need to be more explicit/verbose with "let"! Would students be less inclined to this confusion if we said, for instance, "Suppose P is any polygon, with no other assumptions about its properties"? – Brendan W. Sullivan Feb 24 at 7:02
• Isn't that exactly what "Let P be a polygon" means outside our classroom? @BrendanW.Sullivan – Chris Cunningham Feb 24 at 13:36
• I should say that this confusion arose in written questions, not verbal explanations, where it is easy to elaborate as @BrendanW.Sullivan suggests. – Joseph O'Rourke Feb 24 at 15:05
• @ChrisCunningham Yes, exactly. But it seems like maybe the problem is students don't realize that the single word "let" for us means all of that other stuff. I'm recommending that we spell it out for them a lot. Emphasize that idea in class so that, on a written problem like in OP, part of the question is assessing their understanding that "Let P be a polygon" means to consider all possible polygons. – Brendan W. Sullivan Feb 24 at 16:17
• My initial thought was that "convex" was intended, because omissions such as this are common (sometimes on purpose, sometimes just overlooked), and in my opinion to avoid having a gotcha question you'd want to say something like "polygon (convex or not)", unless prior context or reader exposure was that non-convex polygons arose often enough that your intended audience should be thinking about them (which I can't judge based only on what you've posted). So regarding "how do you circumvent it", I would say by anticipating common/expected assumptions unless you're specifically testing that. – Dave L Renfro Feb 24 at 16:55

Many logicians that I have spoken to have concurred with my assessment that this is an issue of the misleading use of "let". Many teachers use this word in two very different and incompatible ways. The first is universal quantification, as in your example. The second is existential instantiation, as in "Let $$z = \exp(x+y)$$. Then [blah blah] about $$z$$.".

The solution is simple. Do not use the bare word "let" for universal quantification, because it is not even technically the most precise way of expressing the desired meaning in English. Here are some much preferable alternatives:

• Take/consider any polygon $$P$$. ...
• Let $$P$$ be any given polygon. Then ...

The key word is "any", which is needed to precisely convey the universal quantification. If you want to express universal quantification in a single sentence, you can use the following:

• Given any polygon $$P$$, ...
• For every polygon $$P$$, ...

Note that the English word "any" behaves differently from "every". For example, "If for any $$x∈S$$ we have $$f(x)∈T$$, then ..." actually means "For every $$x∈S$$, if $$f(x)∈T$$ then ..." and not "If for every $$x∈S$$ we have $$f(x)∈T$$, then ...". So to minimize confusion it is better to stick with "for every" in single-sentence universal quantification.

I often hear the excuse that people have been using "let" in the confusing way for so long already, and that students have to learn to interpret it anyway. That is an excuse, because we have no reason not to use precise words in our teaching. As you yourself pointed out emphatically, using the word "any" would make things so much clearer. So we should use it! After students have acquired a proper grasp of logical reasoning (and not before that), facilitated by precise teaching, we can then tell students that some textbooks actually use "let" in the confusing manner, and they will have no trouble with it.

Related to this, students understand quantifiers far better when they are explained in terms of game semantics. Similarly for teaching structural induction. These are also perfectly in line with the use of the phrasing "given any ...".

• Agreed. But I would personally lean towards arbitrary rather than any as it is a more mathematically precise word. – Matthew Daly Apr 13 at 19:19
• @MatthewDaly: Yes, you can use "given an arbitrary polygon $P$", but it is not any more mathematically precise than using "given any". This is standard English. If we do not want to use English, there is always "∀". – user21820 Apr 14 at 6:38
• @MatthewDaly: There is also another reason to stick to "any", because it is more concise. Consider: "Given any $a,b∈\mathbb{R}$ with $a≤b$, and any continuous $f : [a,b]→\mathbb{R}$ such that $f(a) ≤ 0 ≤ f(b)$, there is some $c∈[a,b]$ such that $f(c) = 0$. Similarly for theorems involving more universally quantified variables. =) – user21820 Apr 14 at 6:57