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


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?

  • 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

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.


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.


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