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I have seen two versions of the open mapping theorem. I am trying to understand why they are equivalent.

From wikipedia:

If $X$ and $Y$ are Banach spaces and $A : X \rightarrow Y$ is a surjective continuous linear operator, then $A$ is an open map.

From Royden (paraphrased):

Let $X$ and $Y$ are Banach spaces and $T : X \rightarrow Y$ is a continuous linear operator. $T(X)$ is closed as a subspace of $Y$ iff $T$ is an open map.

How are these equivalent?

EDIT:

I've included the releveant portion in Royden. Indeed, he discusses the image as having inherited the subspace topology from $Y$ -- I missed this before the discussion in the comments, thanks!

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  • $\begingroup$ Presumably, you mean $A$ in your second statement, not $T$? $\endgroup$
    – cmk
    Aug 4, 2019 at 14:35
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    $\begingroup$ @FedericoFallucca. If $T$ is open and $T(X)$ is closed, then $T(X)$ must equal $Y$. That we agree. On the other hand, The zero linear map $T:X\to Y$ obviously satisfies that $T(X)$ is closed in $Y$, but clearly $T$ is not open. Doesn't this contradict the theorem? $\endgroup$ Aug 4, 2019 at 15:23
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    $\begingroup$ No because you must specify what map $T$ is open: is it open the map $ T:X\to Y$ or the map $T:X\to T(X)$? $\endgroup$ Aug 4, 2019 at 15:24
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    $\begingroup$ @FedericoFallucca In that case, my friend, the theorem is not formulated precisely. We are given a continuous linear operator $T:X\to Y$. If we say that this operator is open, we mean that $TU$ is open in $Y$ for every $U$ open in $X$. There is no other definition. So if you say "$T(X)$ is closed iff $T$ is open", but you really want to say "$T(X)$ is closed iff $T$ is open as a map between $X$ and the range of $T$ in the relative topology", then you should say it. As it stands, the theorem is simply false, and I don't believe Royden would write such a thing. $\endgroup$ Aug 4, 2019 at 16:04
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    $\begingroup$ Perfect, then it is clear that is not true the equivalenze of two statements, because I get an example in which T is continuos, linear and T(X) is closed but T is not open $\endgroup$ Aug 4, 2019 at 16:14

1 Answer 1

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It is clear that the second result implies the first result.

If it is true the first result, we consider a map $T:X\to Y$ such that $T(X)$ is closed. Each closed subspace of a Banach Space $Y$ is also a Banach Space so $T(X)$ is a Banach Space and $T:X \to T(X)$ is a surjective continuos linear operator between Banach Spaces; so, by first result, $T:X\to T(X)$ will be an open map. In any case, it is not true that $T: X\to Y$ will be an open map because in general $T(X)$ it is not open in $Y$, infact if it is open, then $T(X)$ will be a nonempty open and closed subspace of the connected space $Y$, so $T(X)=Y$. An example can be $T:\mathbb{R}\to \mathbb{R}^2$ such that

$T(x):=(x,0)$ . The map is continuos and Linear while the two spaces are Banach, so $T:\mathbb{R}\to \mathbb{R}\times \{0\}$ is open while

$T:\mathbb{R}\to \mathbb{R}^2$ is not open because $T(\mathbb{R})= \mathbb{R}\times \{0\}$ that it is not open in $\mathbb{R}^2$.

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