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The open mapping theorem states that a bounded surjective linear operator T on a Complete normed space X, mapping it to a Complete normed linear space Y, will be an open map.

My argument is that one can prove this statement without the fact that the spaces are complete!

What follows from this theorem is that a bijective linear operator on complete spaces will have an inverse that is bounded. This I can also prove without needing completeness.

Am I missing a point here? Why is it always required that the spaces are complete, in order for the theorem to hold?

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  • $\begingroup$ And what is your argument? There must be a mistake somewhere, as the theorem fails without completeness, there are counterexamples. $\endgroup$ – Giuseppe Negro May 6 at 15:15
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One direction of the open mapping theorem does not need completeness: if $T:X\rightarrow Y$ should be an open mapping then it is onto. Proving the converse is where completeness comes in, though only for the domain $X$; the theorem holds when $Y$ is only a normed linear space.

During the proof of the converse we need to construct a sequence $\{x_n\}_{n\in {\mathbb N}}$ such that $\|x_n\| \lt \delta ^{n-1}$ where $\delta \gt 1$ is a scaling factor used to control the image of the unit ball $(1-\delta)^{-1}T(B_X)$. We observe that the series $\sum_n x_n$ is therefore absolutely convergent and set its limit to be $x$. Without knowing that $X$ is complete we cannot know that this $x$ exists in $X$. Without that, we cannot be certain that $Y\subset T(X)$ and so that $T$ is open.

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  • $\begingroup$ Thank you for your answer. I do not understand however what you mean with ''onto'' in the second row. Can I ask you to please clarify that? $\endgroup$ – N. Koudsie May 6 at 17:37
  • $\begingroup$ @N.Koudsie Onto means surjective. $\endgroup$ – postmortes May 6 at 17:38
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@Giuseppe Negro Gladly: Say an open set O in X is mapped to TO in Y. I can prove that a ball B(x_0 , l_0) in O will be, with all its elements, mapped to another ball B(Tx_0 , l'_0) in TO, with all of the elements of the new ball being an image of the old ball only. With that, I have proven that there exists a ball around any element of TO, with all its elements still contained in TO, which in turn means that TO is open.

The proof that I can provide is based on continuity and boundedness arguments, which are completely independent of completeness.

The proof goes as follows: If an element y is inside the new ball, then due to surjectivity, it must be an image Tx of some element x in X. Continuity and boundedness will forbid the possibility of the element x being outside the old ball.

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    $\begingroup$ If $X$ isn't complete then you have no guarantee your $x$ exists, regardless of continuity and boundedness. Consider that you can approach $\sqrt{2}$ in ${\mathbb Q}$ as closely as you like with a bounded, continuous function but $\sqrt{2}$ will not ever be rational. $\endgroup$ – postmortes May 6 at 20:02
  • $\begingroup$ @postmortes Thank you. I finally saw what was missing! $\endgroup$ – N. Koudsie May 8 at 9:22
  • $\begingroup$ I disapprove the downvote. Sure, there is a mistake, but the purpose of this answer was exactly to find it. I upvoted. $\endgroup$ – Giuseppe Negro May 17 at 0:15

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