Prove the sum and product of real numbers are continuous functions

Question

To start with, let us study the continuity of the sum of real numbers.

Precisely speaking, let us consider the function $f:\textbf{R}^{2}\to\textbf{R}$ such that $f(x,y) = x + y$.

We want to prove that for every $\varepsilon > 0$, there is a $\delta > 0$ such that \begin{align*} \sqrt{|x - a|^{2} + |y - b|^{2}} < \delta \Rightarrow |x + y - a - b| < \varepsilon \end{align*}

Since $|x - a| \leq \sqrt{|x - a|^{2} + |y - b|^{2}}$ as well as $|y - b| \leq \sqrt{|x - a|^{2} + |y - b|^{2}}$, for every $\varepsilon > 0$, there is $\delta = \varepsilon/2 > 0$ such that \begin{align*} \sqrt{|x - a|^{2} + |y - b|^{2}} < \delta \Rightarrow |x + y - a - b| \leq |x-a| + |y-b| < 2\delta = \varepsilon \end{align*}

and we are done.

Let us now try to prove the continuity of the product.

Precisely speaking, let us consider the function $f:\textbf{R}^{2}\to\textbf{R}$ such that $f(x,y) = xy$.

We want to prove that for every $\varepsilon > 0$ there is a $\delta > 0$ such that \begin{align*} \sqrt{|x - a|^{2} + |y - b|^{2}} < \delta \Rightarrow |xy - ab| < \varepsilon \end{align*}

I do not know how to proceed, but I tried to approach it as follows: \begin{align*} |xy - ab| & = |xy - xb + xb - ab|\\\\ & = |x(y-b) + b(x-a)|\\\\ & \leq |x||y-b| + |b||x-a|\\\\ & \leq |x|\delta + |b|\delta \end{align*}

Could someone finish it or propose another way to solve it?

EDIT

In accordance to the comment of @Noobmathematician, I have tried the following approach. Any comments are appreciated.

Let us suppose the sequence $(a_{n},b_{n})$ converges to $(a,b)$ in $\textbf{R}^{2}$. We shall prove that $f(a_{n},b_{n})$ converges to $f(a,b)$ in $\textbf{R}$. But a sequence converges in $\textbf{R}^{2}$ iff it converges coordinatewise. That is to say, $a_{n}\to a$ and $b_{n}\to b$.

Consequently, $a_{n}$ is bounded: $|a_{n}|\leq M$

Hence we have \begin{align*} |a_{n}b_{n} - ab| & =|a_{n}b_{n} - a_{n}b + a_{n}b - ab|\\\\ & = |a_{n}(b_{n} - b) + b(a_{n} - a)|\\\\ & \leq |a_{n}||b_{n} - b| + |b||a_{n} - a|\\\\ & \leq M|b_{n} - b| + |b||a_{n} - a| \end{align*}

Taking the limit from both sides and due to the squeeze theorem, one concludes that \begin{align*} \lim_{n\rightarrow\infty}|a_{n}b_{n} - ab| = 0 \Longrightarrow \lim_{n\rightarrow\infty}(a_{n}b_{n} - ab) = 0 \Longrightarrow \lim_{n\rightarrow\infty}a_{n}b_{n} = ab \end{align*}

and we are done.

Answer 1

What you did is right and precise . But I would have used sequential criteria for continuity . Trust me sequential criteria will help you in many tricky situations and avoiding $\epsilon-\delta$ mess.

$$(x_n,y_n) \to (a,b) \implies x_n \to a \text{ and }y_n\to b$$ Hence $$(x_n+ y_n)\to( a+ b) \text{ and }(x_n\cdot y_n)\to (a\cdot b)$$

Answer 2

An other way.

You begin by proving that If $ f $ is continuous, then $ f^2 $ is continuous (easy), then

use the identity

$$xy = \frac{(x+y)^2-(x-y)^2}{4}$$ to prove that $$(x,y)\mapsto xy$$ is continuous.