# Equivalence between two versions of Open Mapping Theorem

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!

• Presumably, you mean $A$ in your second statement, not $T$?
– cmk
Aug 4, 2019 at 14:35
• @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? Aug 4, 2019 at 15:23
• No because you must specify what map $T$ is open: is it open the map $T:X\to Y$ or the map $T:X\to T(X)$? Aug 4, 2019 at 15:24
• @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. Aug 4, 2019 at 16:04
• 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 Aug 4, 2019 at 16:14

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 $$\mathbb{R}^2$$.