Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be continuous map . Then which one of the following is true?

(a) $ f(A) $ is bounded for all bounded subsets $ A $ of $\ \mathbb{R} $ .

(b) $ f^{-1}(A) $ is compact for all compact subsets $ A $ of $ \mathbb{R} $.

Answer The function $ f(x)=\frac{1}{x} $ is continuous on $ (0, \infty ) $ . Now $ \ (0,1) $ is bounded. But $ f((0,1))=(1, \infty) $ is unbounded . Hence option $ (a) $ is wrong.

So option $ (b) $ is correct. But I am not sure. Any help is there?

  • 1
    $\begingroup$ The wording of your question makes it seem like $f$ is defined (and continuous) on $\textit{all}$ of $\mathbb{R}$, so your counterexample would not work. $\endgroup$ – pwerth Jun 22 '17 at 22:18
  • $\begingroup$ so which option is true ? $\endgroup$ – M. A. SARKAR Jun 22 '17 at 22:21
  • $\begingroup$ $ f(x)=\tan x $ is a periodic function. Now $ f([0, \frac{\pi}{2} ))=(0, \infty). $ So option (a) is not correct. $\endgroup$ – M. A. SARKAR Jun 22 '17 at 22:24
  • $\begingroup$ The tangent function is not continuous on $\Bbb{R}$, $\endgroup$ – matboy Jun 22 '17 at 22:27
  • $\begingroup$ @matboy: duhhh??? $\endgroup$ – Rob Arthan Jun 22 '17 at 22:29

Your counterexample to $(a)$ is wrong. You need to find a continuous function defined on $\mathbb{R}\to \mathbb{R}$ which is unbounded on bounded sets. However $\frac1x$ is not defined on $0$.

In fact, option $(a)$ is true. As $A$ is bounded, therefore $\overline{A}$ is compact(Heine Borel Theorem). Hence, $f(\overline A) $ is bounded.

Now $ f(A) \subseteq f(\overline A) $. Therefore, $f(A)$ is also bounded.

Option $(b)$ is false.

Counterexample: Let $f(x)=\sin x$. Then $\{0\}$ is compact. However, $f^{-1}\{0\}=\{n\pi:n \in \mathbb{Z}\}$ which is not bounded, hence not compact.


Your example is not adequate because the inverse function is not defined on $\Bbb{R}$.

Proposition $a)$ is true. The closure of $A$ is a compact set, because it is a bounded (since $A$ is) and closed set of $\Bbb{R}$. Since any continuous function on a compact set is bounded, $f$ is bounded on the closure of $A$, and so it's bounded on $A$.

Proposition $b)$ is false. Take the contant function equal to $1$, then $\{1\}$ is compact but $f^{-1}(\{1\})={\Bbb{R}}$ is not.

  • $\begingroup$ is option (a) is true? $\endgroup$ – M. A. SARKAR Jun 22 '17 at 22:29

Hint: Take $f:\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=1$ for all $x \in \mathbb{R}$.


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