We say that a function \(f\) is continuous at a point \(x\) in its domain (or at the point \((x,f(x))\)) if, for any open interval \(S\) containing \(f(x)\text{,}\) there is an open interval \(T\) containing \(x\) such that if \(t\in T\) is in the domain of \(f\text{,}\) then \(f(t)\in S\text{.}\)
Let’s show that this definition of continuity behaves the way we expect from calculus. The first example will make it particularly easy to find \(T\text{,}\) given \(S\text{.}\)
We say that a function \(f\) is sequentially continuous at a point \(x\) if, for every sequence \(\set{x_i}_{i=1}^\infty\) (in the domain of \(f\)) converging to \(x\text{,}\) it is also true that \(\set{f(x_i)}_{i=1}^\infty\) converges to \(f(x)\text{.}\)
So these concepts are equivalent on the real numbers. However, we still distinguish between them, since this isn’t the case for all sets. See, for instance:
The sequential way of thinking of continuity makes proving some basic facts easier. Be clear on the logic in the next two, and don’t forget what we’ve already proved or stated about sequence convergence.
Before finally moving on to calculus proper, we’ll deal briefly with two foundational theorems about \(\R\) and continuous functions that have everything to do with completeness.
Let \(I\) be a closed interval and suppose that \(f\) is a function which is continuous on \(I\text{.}\) Then there exists \(x_M \in I\) such that \(f(x_M) \ge f(x)\) for all \(x\in I\text{.}\)
Suppose \(f\) is a continuous function on a closed interval \(I = [a,b]\text{,}\) and that \(f(a) \lt 0 \lt f(b)\text{.}\) Then there is a point \(x\in [a,b]\) such that \(f(x) = 0\text{.}\)
Suppose \(f\) is a continuous function on a closed interval \(I = [a,b]\) and that \(f(a) \lt c \lt f(b)\text{.}\) Then there is a point \(x\in [a,b]\) such that \(f(x) = c\text{.}\)
Note that this is also true if \(f(b) \lt c \lt f(a)\text{.}\) One way to get a handle on these facts is to find counterexamples when the hypotheses are not satisfied.
Find a continuous function on \(\R\) which does not satisfy the conclusions of the Extreme Value Theorem; do the same for a discontinuous function on \([0,1]\text{.}\)
Find a continuous function on \([0,1]\cup [2,3]\) with \(f(0) = -1\text{,}\)\(f(3) = 1\) which does not satisfy the conclusions of the Intermediate Value Theorem; do the same for a discontinuous function on \([0,1]\) where \(f(0) = -1\text{,}\)\(f(1) = 1\text{.}\)
Proving the EVT is most often done via the Bolzano-Weierstrauss Theorem and the sequential definition of continuity (or other definitions); it’s rather tricky to do it using only the supremum definition. Here is one piece of the proof (this piece actually applies to non-continuous functions).
Let \(f\) be a function on \([0,1]\text{,}\) and assume that the image \(f([0,1])\) has a supremum. Show there is a sequence of points \(\set{x_i}_{i=1}^\infty\) in \([0,1]\) such that \(\set{f(x_i)}_{i=1}^\infty\) converges to that supremum.
Proving the IVT, on the other hand, can definitely be done using just the set definitions of completeness and continuity. The next problem should get your creative juices flowing as to how to find the \(x\) such that the IVT holds for that \(x\text{.}\) Be careful to show that your \(x\) exists!
Let \(f\) be a function on \([0,1]\) such that \(f(0) = -1, f(1) = 1\text{,}\) and \(f([0,1]) = \set{-1,1}\text{.}\) Show (without using the IVT) that there is an \(x\in [0,1]\) such that \(f\) is not continuous at \(x\text{.}\)
The Intermediate and Extreme Value Theorems combine in a deep way. If we had time for just a bit more topology (maybe at the end of the semester?), we could have introduced the words compact and connected; these more advanced concepts are necessary to describe the proper generalization of the following result for more general spaces than \(\R\text{.}\)
Find a function \(f\) on a closed interval \(I\) such that \(f(I)\) is also a closed interval, but for which \(f\) is not continuous for some \(x\in I\text{.}\)
Find a continuous function \(g\) on a closed interval \(I = [a,b]\) such that \(g(a) = g(b)\) but \(g(I)\) is not (just) the single point \(g(a)\text{.}\)
