4
$\begingroup$

Let $f$ be a continuous function from the reals $\mathbb{R}$ onto $I=[0,1]$ with usual topology. Prove that if $C$ is a subset of $I$ and the preimage of $C$ is closed in $\mathbb{R}$ then $C$ is closed in $I$.

My attempt is to use normal space properties but it does not help.

$\endgroup$
  • $\begingroup$ usual topology on $\mathbb R$ & standard topology on $\mathbb R$ same? $\endgroup$ – Styles Aug 8 '16 at 19:36
  • $\begingroup$ Yes, you are right $\endgroup$ – Gob Aug 8 '16 at 19:44
1
$\begingroup$

Hint: it will be easier (in my opinion) to show that $C^c$ must be open in I, take a point in $x \in C^c$, and look for $r>0$ s.t $B(x,r) \subset C^c$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint: There is a compact interval $J=[c,d]$ such that $f(J)=I$. If $(x_n)\subseteq C$ converges in $I$, take preimages and use compactness of $J$ to show that $\lim x_n\in C$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Just I am confused because $f$ is onto and you said there is $J$ such that $f(J)=I$ .So, $J$ must be $\mathbb{R}$. Could you explain more about you hint $\endgroup$ – Gob Aug 8 '16 at 19:35
  • $\begingroup$ @Gob I should've been more explicit: There exists a compact interval $J$ with $f(J)=I$. I'm sorry about that. $\endgroup$ – Luiz Cordeiro Aug 8 '16 at 20:01
0
$\begingroup$

Let $D$ be the complementary space of $C$. We are going to show that $D$ is open. Suppose that $D$ is not open. There exists $y\in D$ such that for every open interval $y\in J$, $J\cap[0,1]\cap C$ is not empty. Since $f$ is continuous, $f^{-1}(J\cap [0,1])$ is open. Let $x\in f^{-1}(J\cap [0,1])$ such that $f(x)=y$, there exists an interval $I_x$, $x\in I_x\subset f^{-1}(J\cap [0,1])$, we choose $I_x=(a,b)$ maximal.

If $a$ and $b$ are not real numbers, $I_x=R$ $D=[0,1]$ and $C$ is the empty subset contradiction, since the empty subset is closed.

Suppose that $a$ and $b$ are real numbers. $f((a,b))$ is an interval contained in $D$. $a$ and $b$ are contained in $f^{-1}(C)$. This implies that $f(a)$ and $f(b)$ cannot be in the interior of $f((a,b))$ since it is contained in $D$. We deduce that $f([a,b])]=[f(a),f(b)]$ or $[f(b),f(a)]$. $y$ cannot be in the interior of $f((a,b))]$, since $y$ is an accumulation point of $C$. This implies that $y=f(a)$ or $y=f(b)$ contrariction.

Suppose that $I_x=(a,+\infty)$, $f(I_x)$ is an interva $[c,d]$ where $0<c<d<1$, the intermediate value theorem implies that $f((-\infty,a))=[0,1]$, again there exists $x'\in (-\infty,a)$ such that $f(x')$. $I_{x'}$ the maximal open interval contained in $f^{-1}(D)$ and containing $x'$ is of the form $(-\infty,b)$. We deduce that $f^{-1}(C)\subset [b,a]$ and is compact since it is closed. This implies that $C$ is closed. Contradiction. We deduce that $f(I_x)$ contains $0$ or $1$. Suppose that $0\in f(I_x)$, since $y$ is an accumulation point of $C$, $f(I_x)=[0,y]$, $f(a)=0$ since $f(a)\in C, f(a)\neq y $ and if $f(a)$ in the interior of $[0,y]$ the intermediate value theorem implies there exists $x'\in I_x$ such that $f(x')=f(a)$. Since $f$ is surjective, $f((-\infty, a))=[0,1]$ (use the intermediate value theorem). There exists $x_1$ in $(-\infty, a)$ such that $f(x_1)=y$. Again we apply the previous step to show that the maximal open interval in $f^{-1}(D)$ containing $x_1$ is of the form $(-\infty,b)$. This implies that $f^{-1}(C)$ is contained in $[b,a]$, since it is a closed subset we deduce that $f^{-1}(C)$ is compact. This implies that $C$ is compact since the image of a compact subset by a continuous map is compact and henceforth closed. Contradiction. The case where $1\in f(I_x)$ is similar.

The case where $I_x=(-\infty,a)$ is similar.

| cite | improve this answer | |
$\endgroup$

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.