4
$\begingroup$

Let $I_1,I_2,I_3$ be intervals $\subset \mathbb{R}$. Suppose $f:I_1 \to I_2$ is a surjective continuous function and $g: I_2 \to I_3$ is a discontinuous function. Must the composition $g \circ f$ be discontinuous?

There are some easy counter-examples if $f$ is not assumed to be surjective, e.g. taking $f$ to be a constant function, or in a way that "dodges" the discontinuous point(s) of $g$.

However if such "dodging" is prohibited, I fail to construct such functions nor find an answer from many similar questions on this site. So I am interested to know whether counter-examples exist? If not, is there a proof? Does it have something to do with intermediate value theorem?

$\endgroup$

3 Answers 3

1
$\begingroup$

Here is a proof without compactness of the intervals.

Suppose that $f$ and $g\circ f$ are continuous, $f$ is surjective. I want to show that $g$ is continuous.

Take $x\in I_2$ and a sequence $(x_n)$ in $I_2$ with $x_n\to x$. Since $I_2$ is an interval, there are $x'<x''$ in $I_2$ such that $x\in [x',x'']\subset I_2$ and $x_n\in[x',x'']$ for all $n$. (If this would not be possible, then $I_2$ would be a singleton.)

Since $f$ is surjective, there are $y,y',y''$ in $I_1$ such that $f(y)=x$, $f(y')=x''$, $f(y'')=x''$. Define the interval $J:=[\min(y,y',y''),\ \max(y,y',y'')]$. Due to the intermediate value theorem $f(J) \supset [x',x'']$.

Now for every $n$ there is $y_n\in J$ with $f(y_n)=x_n$. Since $J$ is compact, there is a converging subsequence $(y_{n_k})$ with limit $z$. By continuity of $f$, $f(z)=x$. Then by continuity of $g\circ f$ $$ g(x_{n_k}) = g(f(y_{n_k})) \to g(f(z)) = g(x). $$ Now, we can repeat this argument for each subsequence of $(x_n)$. The limit $g(x)$ does not depend on the chosen subsequence, so $g(x_n)\to g(x)$, and $g$ is continuous.


Can this proof be generalized to higher dimensions?

$\endgroup$
1
$\begingroup$

If $g$ is discontinuous at $x=a$, then $gf$ cannot be continuous. Suppose $gf$ is continuous. For any sequence $\{x_n\}\in I_2, x_n\to a$, due to the surjectivity of $f$, $\exists \{y_n\}\in I_1, f(y_n)=x_n,\forall n$.

$y_n\in I_2$ which is compact, so there must be a subsequesce $y_{n_k}$ that converges. Let $\lim_{k\to \infty}y_{n_k}=L$.

Since $f$ is continuous, $f(L)=\lim_{k\to \infty} f(y_{n_k})=\lim_{k\to \infty}x_{n_k}=\lim x_n=a$. Also, $gf(L)=g(a)$.

So, $\lim g(x_n)=\lim gf(y_{n_k})=gf(L)=g(a)$, thanks to the continuity of $gf$.

The result is that we establish the continuity of $g$ from the continuity of $f$ and $gf$. So $gf$ cannot be continuous.

$\endgroup$
3
  • $\begingroup$ $y_n$ has to be in $I_1$. Also compactness of $I_1$ is required. $\endgroup$
    – daw
    May 23, 2019 at 7:47
  • $\begingroup$ @daw But everybody knows it, so no need to write that down.... $\endgroup$
    – Ma Joad
    May 23, 2019 at 10:25
  • $\begingroup$ @daw Ok I will add it. Please tell me anything else that is inaccurate in my answer. $\endgroup$
    – Ma Joad
    May 23, 2019 at 14:38
-1
$\begingroup$

Given that $g$ is discontinuous there exists $x_0 \in I_2$ and a sequence $(x_n)_{n \in \mathbb{N}} \subseteq I_2$ such that $\lim\limits_{n \to \infty}x_n = x_0$ and $g(x_0) \neq \lim\limits_{n \to \infty}g(x_n)$. Now let $y_0 \in f^{-1}(x_0)$ and for any $n \in \mathbb{N}$ let $y_n \in f^{-1}(x_n)$ be given. It follows \begin{align*} (g \circ f)(y) = g(x_0) \neq \lim\limits_{n \to \infty} g(x_n) = \lim\limits_{n \to \infty} (g \circ f)(y_n). \end{align*} We have therefore shown that $g\circ f$ is discontinuous at $y$.

$\endgroup$
3
  • $\begingroup$ Who says $\lim_{n\to\infty}y_n$ has anything to do with $y$ (or $y_0$, you used both names for the same variable)? $\endgroup$
    – Ingix
    May 22, 2019 at 12:51
  • $\begingroup$ Without showing that $\lim_{n\to\infty} y_n=y_0$, this doesn't prove that $g\circ f$ is not continuous. $\endgroup$
    – 5xum
    May 22, 2019 at 13:07
  • $\begingroup$ I guess @neca has choosen the preimages wiseley and just forgot to tell us. $\endgroup$
    – Jupp
    May 22, 2019 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.