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I am trying to read a proof of Dini's theorem and the proof says that the set $E_n=\{x \mid g_n(x)<\varepsilon\}$ is open because $g_n$ is continous and $E_n=g_n^{-1}(-\infty,\varepsilon)$, why does that mean that $E_n$ is open? I would like to see a rigorous proof.

Viktor Vaughn
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איתן כחלון
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    A function is continuous if and only if the preimage of open sets is open. – Andrew Jun 21 '23 at 18:51
  • But can you prove this theorem? – איתן כחלון Jun 21 '23 at 18:52
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    It’s straightforward to prove and you should do it as an exercise. Alternatively, a proof can be found in any analysis text. See chapter 4 of Rudin’s PMA, for example. – Andrew Jun 21 '23 at 18:54
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    For many people, what @AndrewZhang writes is the definition of continuity. If your definition of continuity is different, write it down. – Torsten Schoeneberg Jun 21 '23 at 18:54
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    A function is continuous if and only if the preimage of every open set is open. $(-\infty, \varepsilon)$ is open and $g_n$ is continuous. Thus $g_n^{-1}(-\infty, \varepsilon)$ is open. Which of these steps is unclear? – Viktor Vaughn Jun 21 '23 at 18:56

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The definition of a function being continuous is exactly that preimages of open sets are open. Maybe you have learnt another equivalent definition, but this one is the most general one (as it also works for topological spaces). On Wikipedia you can find all the equivalent formulations and their proof of equivalence.

JakobGFF
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