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Given that for each $\delta>0$ you can find $x$ s.t $\|x\|=1$ $\|T(x)\|<\delta$ show $T$ cannot be an open map.

The claim was made here: When is the image of a linear operator closed?

I am not quite sure how this contradicts $T$ being an open map. It just says that image of an openball around $0$ of radius larger than $1$ is an open set, that contains an open ball around $0$ which is true for an open map as $T(0)=0$ What am I missing here?

Sorfosh
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2 Answers2

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My issue with the answer in the linked post is more the purpose of that answer. The claim is, $T$ open $\implies$ the $\text{im}(T)$ is closed. Isn't this trivial by the definition of an open map? If $T: X\to Y$ is open then it is surjective, so it covers $Y$. Since $Y$ is both open and closed in the $\|\cdot\|_{Y}$ topolgy then the $\text{im}(T)$ is open and closed.

So I thought about this question a little more and seem to have come to the conclusion that the assumption made is not sufficient to conclude $T$ is not open. The assumptions made only tell us how far from the origin the image of specific singletons are and does not imply anything about the topology of the image of any neighbourhood $B_{X}(0;\varepsilon)$.

If I am incorrect with this I would be very interested to know how one can conclude the desired result under the given assumptions.

Zeta-Squared
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  • Ah so it is a single $x$ that does it. I thought for each $\delta$ there is such $x$. That would explain it. Yes, your issue with the linked post is correct, it was also raised in the comments of the question. – Sorfosh Aug 09 '20 at 01:34
  • @Sorfosh Ah I just read the comments. You're right. – Zeta-Squared Aug 09 '20 at 01:36
  • @Sorfosh Sorry I believe I have messed up in my reasoning, now that I re-read my post. Non-trivial kernel implies not injective. Apologies. You should probably remove the answer till this is rectified. – Zeta-Squared Aug 09 '20 at 01:38
  • ah you are right :P I got fooled too. – Sorfosh Aug 09 '20 at 01:39
  • @Sorfosh I don't think the assumption gives the desired conclusion – Zeta-Squared Aug 09 '20 at 05:53
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This is just false, as seen by thinking of any open linear map with non-trivial kernel. For instance, the hypothesis is obviously true for the projection $\mathbb{R}^2\rightarrow \mathbb{R}$, but this map is open. You may find more examples here.

The first "Thrm" in the linked post is true but for simpler reasons. As noted above, what's true is that an open map between normed spaces is automatically surjective.

I think what the author of the answer was thinking is that "If $T$ is open then $T$ is bounded below" and they've appealed to the contrapositive of that (incorrect) statement. What is true is that if $T:X\rightarrow Y$, for $X$ and $Y$ normed space, is open and injective then $T$ is bounded below. The reason is that $T(\{x\mid||x||=1\})\subset Y\setminus T(B(0,1/2))$ by injectivity and $Y\setminus T(B(0,1/2))\subset Y\setminus B(0,\delta)$ for some $\delta>0$ since $B(0,\delta)\subset T(B(0,1/2))$ by openness.

user293794
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  • This is great. Thank you, I couldn't quite put my finger on what was going wrong the statement presented, even though my feeling was that it isn't true. – Zeta-Squared Aug 11 '20 at 01:52
  • I spent some time trying to prove it too even though it felt wrong until I considered that it might be wrong at which point it was easy. So I guess the lesson is don't trust everything you read on the internet :p – user293794 Aug 11 '20 at 13:10