# Prove set is open given continuous function

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?

• Remind that union of open subsets is open whereas union of closed subset is not necessarily closed. – Blumer Dec 9 '18 at 18:46
• 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. – Melody Dec 9 '18 at 19:16

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.