This is an ongoing series intended to help aspiring young (or young at heart) learners make the transition from the computation-based mathematics taught in US public schools and proof-based mathematics found at higher levels of mathematics education. See here for Part One. For a list of notation and definitions, see here.
Proofs make math go. When it comes to verifying whether or not a statement is true, mathematicians have the highest possible bar: a deductive proof. A deductive proof of a statement from assumptions is such that the statement is always true if the assumptions are true. Contrast this with something like the scientific method, which argues that a statement is likely true or false based on observational tests. In this case, it’s always possible (albeit unlikely with proper controls) that our observations were affected by some unconsidered factor, or random chance gave us poor observations to work from. In math, we do not allow arguments that could fall victim o this. In some sense, we account for every possible factor, even the most zany of possibilities, to ensure that the statement is always true.
At a basic level, definitions make proofs go. Definitions are exactly what you think they are based on the usual use of the word, but with an important caveat to consider. In everyday language, we have an intuition about certain words, phrases, and terms that may occasionally run counter to the “official” definition. Even worse (from a mathematician’s perspective,) some definitions depend on context or a person’s subjective taste: A tasty meal for me might not be a tasty meal for you. In mathematics, this subjectivity cannot happen. If your definition is ambiguous, then it is not a mathematical definition at all. This idea of being free from ambiguity is sometimes known as being “well-defined.” We’ll demonstrate how this might lead to struggles with your intuition with a familiar term from calculus, but first we need some background.
Definition: A topological space is a set \(X\) together with a set \(T\subseteq\mathcal{P}(X)\) satisfying the following properties:
- \(\emptyset\in T\) and \(X\in T\)
- If \(A,B\in T\), then \(A\cap B\in T\)
- If \(Y\subseteq T\), then \(\bigcup_{A\in Y} A\in T\)
We call \(X\) the space, \(T\) the topology, and the elements of \(T\) open sets.
Definition: Let \(X\) and \(Y\) be topological spaces with topologies \(T_X\) and \(T_Y\) respectively. A function \(f:X\to Y\) is continuous if \(f^{-1}(A)\in T_X\) for all \(A\in T_Y\).
Now consider the function \(f:\mathbb{N}\to\mathbb{R}\) defined by \(f(x)=\pi+x\). Let the topology for \(\mathbb{N}\) be \(\mathcal{P}(\mathbb{N})\). Now notice that no matter what the topology on \(\mathbb{R}\) is, \(f\) is a continuous function: If \(Y\subseteq\mathbb{R}\), then \(f^{-1}(A)\subseteq\mathbb{N}\). Therefore by definition, \(f^{-1}(A)\in\mathcal{P}(\mathbb{N})\). Thus the function is continuous by our definition of continuity, as the inverse of any subset (and hence all open subsets) is open. If we were to graph this function though, our intuition might tell us that it isn’t continuous as it is just a handful of discrete points in the plane. Strange situations like these are why paying attention to definitions is so important in mathematics.
Congratulations! You’ve already seen a proof now. We proved that the function \(f\) is continuous using the definition. Let’s go through it again, changing the format slightly to make it a bit more clear what is going on. First, we want to prove that \(f\) is continuous. By the definition of continuity, that means that we need to show that \(f^{_1}(A)\in\mathcal{P}(\mathbb{N})\) for all \(A\in T_\mathbb{R}\). (Here \(T_\mathbb{R}\subseteq\mathcal{P}(\mathbb{R})\). Normally the continuity of a function would depend on the topology of its image, but in this case the topology of the domain is nice.) Of course it is true, by the definition of the inverse image, that \(f^{-1}(X)\subseteq\mathbb{N}\) for all \(X\subseteq\mathbb{R}\), so \(f^{-1}(X)\in\mathcal{P}(\mathbb{N})\). In particular, this is true for all \(X\in T_\mathbb{R}\). Thus \(f\) is continuous.
Notice that at each step, we proceeded in the simplest way we could: we wanted to show \(f\) was continuous, so we attempted to show directly that it satisfied the definition. To show that it satisfied the definition, we looked directly at the inverse image. Using that definition and the definition of the power set, we were able to finish our argument. This style of arguing, just looking at definitions and moving in the simplest possible way forward, is sometimes known as “following your nose.” Following your nose isn’t always going to work out, though, so next time we’ll go over some common proof strategies.