Proof of continuity using sequential definition of continuity

Question

I have seen proofs using the delta-epsilon definition of continuity, and they make perfect sense, but I have not found one proof using the sequential definition of continuity.

For example, when given functions, $f$ and $g$ that are continuous on [$a,b$], prove that the function $h=f+g$ is also continuous on [$a,b$]. I also have not seen a proof of $fg$ being continuous on [$a,b]$.

Does a proof exist that does not use the delta-epsilon definition? If so, is it a less concrete proof when using the sequential definition?

Answer 1

Yes.

Let $f$, $g \colon D \to \mathbb{R}$, where $D$ is some nonempty subset of $\mathbb{R}$. Let $a \in D$. Suppose both of $f$ and $g$ are continuous at $a$. We can show that $f+g$ is continuous at $a$ using the sequential criterion.

Let $(x_n)$ be a sequence in $D$ such that $\lim x_n = a$. Since $f$ and $g$ are continuous at $a$, $\lim f(x_n) = f(a)$ and $\lim g(x_n) = g(a)$. Then $$ \lim {}(f+g)(x_n) = \lim {}[f(x_n) + g(x_n)] = \lim f(x_n) + \lim g(x_n) = f(a) + g(a) = (f+g)(a). $$ Since $(x_n)$ was arbitrary, we have shown that for every sequence $(x_n)$ in $D$ that converges to $a$, the sequence $( (f+g)(x_n) )$ converges to $(f+g)(a)$. Therefore, $f+g$ is continuous at $a$.

A similar argument shows that $fg$ is continuous at $a$ (replace the sums above with products).

Answer 2

You could certainly use the sequence definition of continuity, if you are already willing to accept the sum and product rule for limits of sequences. For instance,

Let $f$ and $g$ be continuous at $a$ and let $a_n\to a$. Then $$\lim_{n\to\infty}f(a_n)g(a_n)=\lim_{n\to\infty}f(a_n)\lim_{n\to\infty}g(a_n)=f(a)g(a).$$ Thus, $fg$ is continuous at $a$. You can similarly show $f+g$ is continuous at $a$.

The reason it is not typically done this way is that the $\delta$-$\epsilon$ proof of continuity of products of continuous functions is almost identical to the $\epsilon$-$N$ proof that the limit of a product of sequences is the products of the limits. One of my professors likes to refer to this sort of thing as "conservation of difficulty". Sure, we can rely on the sequence results, but all that does is hide the difficulty.

Answer 3

Yes a proof using the sequential definition exists. In my opinion the reason that most textbooks do not prove the results using the sequential definition is because usually the definition of convergence of a sequence comes after the concept of limit. If the definition of convergence of a sequence comes first, then the continuity of functions would be defined using the sequential definition (the two definitions of continuity are equivalent). In this case a proof of $f+g$ and $fg$ being continuous when $f,g$ are becomes: if $f,g$ are continuous, if $c$ lies in the domain of $f$ and $g$, and if $x_{n} \to c$, then $(f+g)(x_{n}) = f(x_{n}) + g(x_{n}) \to 2c$ and $(fg)(x_{n}) = f(x_{n})g(x_{n}) \to c^{2}$. But you have also to prove first that the limit of the sum (product) of two convergent sequences is the sum (product) of the limits of the convergent sequences.

No such a proof is not less concrete. The sequential definition works because prior to it we have the result that the limit of the sum (product) of two convergent sequences is the sum (product) of the limits of the sequences. To prove this result we have to employ an epsilon-analysis. So if by "concrete" you mean the presence of epsilon, then my conclusion is such a proof is not less concrete.