# I do not understand the last process of proving that $f$ is continuous iff $f^{-1}(G)$ is open.

The problem is:

Let $f$ be a finite function on $\mathbb{R}^n$. show that $f$ is continuous on $\mathbb{R}^n$ if and only if $f^{-1}(G)$ is open for every open $G$ in $\mathbb{R}^1$, or if and only if $f^{-1}(F)$ is closed for every closed $F$ in $\mathbb{R}^1$.

$\left(\Rightarrow\right)$

1. If $f$ is continuous, $f^{-1}(G) \ne \emptyset$.
2. For $\mathbf{x_0}\in f^{-1}(G)$, $f(\mathbf{x_0}) \in G$.
3. Since $G$ is open, there exists $\varepsilon\gt0$ such that $$\left(f(\mathbf{x_0})-\varepsilon, f(\mathbf{x_0})+\varepsilon\right)\subset{}G$$
4. Since $f$ is continuous, for the above $\varepsilon$, there exists $\delta\gt0$ such that $$-\varepsilon \lt f(\mathbf{x})-f(\mathbf{x_0})\lt\varepsilon$$ when $\mathbf{x}\in \left( \mathbf{x_0}-\delta , \mathbf{x_0} +\delta\right).$
5. Thus, $\left(\mathbf{x_0}-\delta, \mathbf{x_0}+\delta\right) \subset f^{-1}(G)$ when $\mathbf{x} \in \left(\mathbf{x_0}-\delta, \mathbf{x_0}+\delta\right)$ and $f(\mathbf{x})\in G$.
6. This means $\mathbf{x_0}$ is the interior point of $f^{-1}(G)$, so $f^{-1}(G)$ is open.

$\left(\Leftarrow\right)$

1. Let $G_0=\left(f(\mathbf{x_0})-\varepsilon, f(\mathbf{x_0})+\varepsilon\right)$ for $\mathbf{x_0}$ and $\varepsilon\gt0$, which is open.
2. Since $f^{-1}(G_0)$ is open by the hypothesis, there exists $\delta\gt0$ such that $$\left(\mathbf{x_0}-\delta, \mathbf{x_0}+\delta\right)\subset f^{-1}(G_0)$$
3. Thus, $f(\mathbf{x})\in\left(f(\mathbf{x_0})-\varepsilon, f(\mathbf{x_0})+\varepsilon\right)$ when $\mathbf{x}\in \left(\mathbf{x_0}-\delta, \mathbf{x_0}+\delta\right)$.
4. This means $f$ is continuous at $\mathbf{x_0}$.
5. Thus $f$ is continuous.

I do not understand the process 11. I think it is needed to prove that both $\displaystyle\lim_{\mathbf{x}\to\mathbf{x_0}}f(\mathbf{x})$ and $f(\mathbf{x_0})$ exist with finite value and they are equal. How did the process 11 come?

• 9. means exactly that $\lim_{x\to x_0} f(x) = f(x_0)$, and they are finite by f being finite. – james1395 May 1 '16 at 15:01
• Uhh..... are they same? – Danny_Kim May 1 '16 at 15:03
• I suppose that $(x_0 - \delta, x_0 + \delta)$ is a notation for $B(x_0, \delta)$? It is a bad notation though – user258700 May 1 '16 at 15:04
• @AhmedHussein Yes, right. I agree with your word. $B\left(\mathbf{x_0}; \delta\right)$ is better than my notation. Thank you. – Danny_Kim May 1 '16 at 15:06
• Step number 1 is wrong (but completely useless). – egreg May 1 '16 at 15:09