0
$\begingroup$

I understand this set is open, but I don't understand the given proof. Can anyone shed some light?

Problem:

Let $R\rightarrow R$ be continuous, show that $\{x:f(x)>0\}$ is an open subset of $R$.

Given Solution:

Suppose $f:R\rightarrow R$ is continuous. Then for some $x\in\{x:f(x)>0\}, \, f(x)=r>0$. To show that $\{x:f(x)>0\}$ is open, consider the open ball $B_{r/4}(f(x))=(f(x)-\frac{r}{4},f(x)+\frac{r}{4})=(\frac{3r}{4},\frac{5r}{4})$. Because $f$ is continuous, the inverse image of any open set is open. In particular, $f^{-1}(B_{r/4}(f(x)))$ is open and contains $x$. Therefore $B_\delta(x)\subset f^{-1}(B_{r/4}(f(x)))$ for some $\delta >0$. Notice however that this implies that $f(B_\delta(x))\subset B_{r/4}(f(x))=(\frac{3r}{4},\frac{5r}{4})$. Hence if $y \in B_\delta(x)$, $f(y)>\frac{3r}{4}>0$, and we see that $B_\delta(x) \subset \{x:f(x)>0\}$ which proves the this set is open.

What I don't get: We consider an open ball, $B_{r/4}(f(x))$. We show its inverse image is open because $f$ is continuous. By the same logic, can't we consider a closed ball of the same radius around x, and then wouldn't its inverse image be closed? Then $f(y)\geq \frac{3r}{4}$ which is a subset of $\{x:f(x)>0\}$.

Isn't it just because the set $\{x:f(x)>0\}$ can't contain any limit points, since if $\{z_n\}_0^\infty \rightarrow x_0,$ such that $f(x_0)=0$, then $\{z_n\}_0^\infty \in$ $\{x:f(x)>0\}$, but not $x_0$, so it doesn't contain its limit points so it can't be closed.

I don't understand the given proof, but I get that it can't contain its limit points so it can't be closed. What am I missing from the given proof?

$\endgroup$
  • 1
    $\begingroup$ Remind that union of open subsets is open whereas union of closed subset is not necessarily closed. $\endgroup$ – Blumer Dec 9 '18 at 18:46
  • $\begingroup$ It is probably more proper to say $x\in\{f(x):f(x)>0\},$ for you want to show it contains a closed ball about each of its points. Also, the set can be empty, so you don't know you can choose an $x.$ Something to not is that $B_{r/4}(f(x))$ is open, buts its intersection with $\text{Image}(f)$ may not be open in the usual topology. However, it is open in the subspace topology. $\endgroup$ – Melody Dec 9 '18 at 19:16
1
$\begingroup$

An equivalent condition for a continuous function is that the inverse image of any open set is open.

Now $(0,\infty)\subset\mathbb R$ is open. Hence $\{x\mid f(x)\gt0\}=f^{-1}(0,\infty)$ is open.

$\endgroup$
0
$\begingroup$

The set $\{x:f(x)>0\}$ can contain limit points. In fact, it can be closed, though in that case we have two choices, it must be $\emptyset$ or $\mathbb{R}$. Consider $f_1,f_2:\mathbb{R}\to\mathbb{R}$ by $f_1(x)=-1$ and $f_2(x)=1.$ Then $$\{x:f_1(x)>0\}=\emptyset,$$$$\{x:f_2(x)>0\}=\mathbb{R}.$$

You are correct that you could apply the same argument in the proof to the closed ball $[f(x)-r/4,f(x)+r/4].$ This doesn't tells us that $\{x:f(x)>0\}$ is closed, for consider that $(-1,1)$ contains a closed ball around every point in its interior, yet is not closed.

The reason to consider an open ball, is because an open set in $\mathbb{R}$ under the usual topology is precisely a set which contains an open ball around each of its points. That is, $U\subseteq\mathbb{R}$ is open if for all $x\in U$ the exists $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq U.$ So if we can show that $\{x:f(x)>0\}$ contains an open ball around each of its interior points, then its open.

Now, a curious thing is that $\{f(x):f(x)>0\}$ is already an open set in the subspace topology, so consider $f^{-1}(\{f(x):f(x)>0\})=\{x:f(x)>0\}$ is open.

$\endgroup$

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.