For students in a first course in analysis or topology, proving that certain function are continuous can be very tricky. However, some proofs which are difficult for students to prove using the $\epsilon-\delta$ definition of continuity are much easier to prove using the topological definition that the pre-image of every open set be open. For example, it is much easier for students to prove that $f(x)=x^2$ is continuous using open sets rather than $\epsilon-\delta$. One particularly challenging proof is showing that multiplication $\cdot \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is continuous. Is anyone aware of a slick way to prove multiplication is continuous using open sets? Any attempt I make seems to more or less be just as complex as the $\epsilon-\delta$ definition.
-
$\begingroup$ Well, how do you define the open sets in $\Bbb R$, or $\Bbb R \times \Bbb R$? $\endgroup$– user98602Commented Feb 5, 2016 at 22:03
-
$\begingroup$ $U$ is open $X$if for every $x\in U$ there is a ball entirely contained in $U$ $\endgroup$– SoraCommented Feb 5, 2016 at 22:07
-
$\begingroup$ Sure, but how do you define "ball"? With an $\varepsilon$. My point is that at some point you need to use the actual topology of $\Bbb R$, defined in terms of $\varepsilon$-balls. $\endgroup$– user98602Commented Feb 5, 2016 at 22:10
3 Answers
Not sure if you would be interested in this approach, but it is easy to prove with the sequence definition of continuity. Suppose $x_n \to x$ and $y_n \to y$, then $$ x_ny_n = (x + (x_n-x))(y + (y_n-y)) = xy + x(y_n-y) + y(x_n-x) + (x_n-x)(y_n-y) $$ so that by triangle inequality $$ |x_ny_n - xy| \leq |x| |y_n-y| + |y||x_n-x| + |x_n-x||y_n-y| $$ since $x_n \to x, y_n \to y$ all three terms on the right tend to zero.
-
$\begingroup$ Ah, that is pretty nice. I'll tuck that away for safe keeping... thanks! $\endgroup$– SoraCommented Feb 5, 2016 at 22:08
-
-
-
$\begingroup$ And inversion of non-zero elements and complex numbers. $\endgroup$ Commented Jul 7, 2017 at 14:16
-
$\begingroup$ @TomCollinge Can you explain how you proved continuity of inversion of complex numbers with this method? $\endgroup$– Žan GradCommented Apr 6, 2018 at 16:24
Or you could (i) show (e,g by picture) that topologically speaking open balls are equivalent to open squares,
(ii) then use the fact that $f(x,y)= xy $ is monotone in each variable $x,y>0$ and
(iii) use the multiplicative properties of $f$ to show e,g that if $.9 x_0<x<1.1 x_0$ (and likewise for $y$),
then $ .9^2 f(x_0, y_0) <f(x,y)< 1.1^2 f(x_0, y_0)$
In other words, make use of the multiplicative structure of the reals as much as possible. This proves continuity in the open first quadrant.
As nullUser said, if we take $x_n\to x$ and $y_n\to y$, we need to show that $x_n y_n \to xy$. However, the triangle inequality part can be a little bit simpler. Notice that $|x_n y_n - xy| = |x_n y_n - xy_n + xy_n - xy|$ $\leq |x_n y_n - xy_n| + |xy_n - xy| = |y_n||x_n-x| + |x||y_n-y|$.
Now, the right hand side goes to zero per assumption, meaning the product also converges.
-
-
$\begingroup$ This is a circular argument -- you're trying to prove that multiplication is continuous, but at the end there you used the fact that multiplication is continuous to show that the right hand side goes to zero. $\endgroup$ Commented Jan 23, 2023 at 13:17
-
$\begingroup$ Doesn't NullUser also use circular reasoning then? In saying |x_n - x| |y_n - y| approaches zero? $\endgroup$ Commented Dec 14, 2023 at 2:43