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  1. How would you proceed in the below, or have I led myself to a dead-end?

    • Let $O$ be an open set from $\mathbb{R}$. $O \subset \mathbb{R}$
    • Let $D \subseteq \mathbb{R}$, be the domain of $f$. i.e. $f : D \rightarrow \mathbb{R}$. For notation ease.
    • Then $f^{-1}(O) = \left\{ d \in D | f(d) \in O \right\}$
    • $f^{-1}(O)$ is open. Proof.
    • The above implies: $\forall x \in f^{-1}(O)$, there is a $r > 0$ s.t. $\left\{ y \in \mathbb{R} | |y-x| < r \right\} \subset f^{-1}(O)$

Now the above looks very similar to the $\epsilon-\delta$ definition of the continuity, but the statement is about the domain of function $f$. So I am stuck.

  1. How come there is a counter-example to this proof (I must be understanding it incorrectly)?

I know there is this question. But I am asking about my steps in 1.

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  • $\begingroup$ If $f:\Bbb R\to\Bbb R$ is a function, then its domain is $\Bbb R$. $\endgroup$ – Lord Shark the Unknown Apr 22 at 11:56
  • $\begingroup$ I wanted to write $d \in D$ in the definition of $f^{-1}(O)$ for better clarity. $\endgroup$ – i squared - Keep it Real Apr 22 at 11:58
  • $\begingroup$ There is no counter-example, there are examples of continuous functions that are not open. $\endgroup$ – Peter Melech Apr 22 at 12:00
  • $\begingroup$ @PeterMelech got it, thanks $\endgroup$ – i squared - Keep it Real Apr 22 at 12:00
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    $\begingroup$ You seem to be confusing open functions with continuous functions. $f$ is continuous iff the preimage of open sets is open, and $f$ is open iff the image of open sets is open. The counterexample shows that $f$ continuous doesn't imply $f$ open. $\endgroup$ – Yagger Apr 22 at 13:23

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