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 $\mathbb{R}^1$.


  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.


  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?

  • 1
    $\begingroup$ 9. means exactly that $\lim_{x\to x_0} f(x) = f(x_0)$, and they are finite by f being finite. $\endgroup$ – james1395 May 1 '16 at 15:01
  • $\begingroup$ Uhh..... are they same? $\endgroup$ – Danny_Kim May 1 '16 at 15:03
  • $\begingroup$ I suppose that $(x_0 - \delta, x_0 + \delta)$ is a notation for $B(x_0, \delta)$? It is a bad notation though $\endgroup$ – user258700 May 1 '16 at 15:04
  • $\begingroup$ @AhmedHussein Yes, right. I agree with your word. $B\left(\mathbf{x_0}; \delta\right)$ is better than my notation. Thank you. $\endgroup$ – Danny_Kim May 1 '16 at 15:06
  • 3
    $\begingroup$ Step number 1 is wrong (but completely useless). $\endgroup$ – egreg May 1 '16 at 15:09

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